The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview...
Transcript of The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview...
![Page 1: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/1.jpg)
The integrating factor method (Sect. 2.1).
I Overview of differential equations.
I Linear Ordinary Differential Equations.I The integrating factor method.
I Constant coefficients.I The Initial Value Problem.I Variable coefficients.
Read:
I The direction field. Example 2 in Section 1.1 in the Textbook.
I See direction field plotters in Internet. For example, see:http://math.rice.edu/ dfield/dfpp.htmlThis link is given in our class webpage.
![Page 2: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/2.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 3: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/3.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 4: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/4.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 5: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/5.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 6: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/6.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 7: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/7.jpg)
Overview of differential equations.
DefinitionA differential equation is an equation, where the unknown is afunction, and both the function and its derivative appear in theequation.
Remark: There are two main types of differential equations:
I Ordinary Differential Equations (ODE): Derivatives withrespect to only one variable appear in the equation.
Example:Newton’s second law of motion: m a = F.
I Partial differential Equations (PDE): Partial derivatives of twoor more variables appear in the equation.
Example:The wave equation for sound propagation in air.
![Page 8: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/8.jpg)
Overview of differential equations.
Example
Newton’s second law of motion is an ODE: The unknown is x(t),the particle position as function of time t and the equation is
d2
dt2x(t) =
1
mF(t, x(t)),
with m the particle mass and F the force acting on the particle.
Example
The wave equation is a PDE: The unknown is u(t, x), a functionthat depends on two variables, and the equation is
∂2
∂t2u(t, x) = v2 ∂2
∂x2u(t, x),
with v the wave speed. Sound propagation in air is described by awave equation, where u represents the air pressure.
![Page 9: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/9.jpg)
Overview of differential equations.
Example
Newton’s second law of motion is an ODE: The unknown is x(t),the particle position as function of time t and the equation is
d2
dt2x(t) =
1
mF(t, x(t)),
with m the particle mass and F the force acting on the particle.
Example
The wave equation is a PDE: The unknown is u(t, x), a functionthat depends on two variables, and the equation is
∂2
∂t2u(t, x) = v2 ∂2
∂x2u(t, x),
with v the wave speed. Sound propagation in air is described by awave equation, where u represents the air pressure.
![Page 10: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/10.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 11: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/11.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:
I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 12: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/12.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)
I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 13: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/13.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 14: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/14.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:
I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 15: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/15.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 16: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/16.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:
I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 17: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/17.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 18: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/18.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:
I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 19: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/19.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 20: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/20.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:
I The equations of QED. (PDE).
![Page 21: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/21.jpg)
Overview of differential equations.
Remark: Differential equations are a central part in a physicaldescription of nature:
I Classical Mechanics:I Newton’s second law of motion. (ODE)I Lagrange’s equations. (ODE)
I Electromagnetism:I Maxwell’s equations. (PDE)
I Quantum Mechanics:I Schrodinger’s equation. (PDE)
I General Relativity:I Einstein equation. (PDE)
I Quantum Electrodynamics:I The equations of QED. (PDE).
![Page 22: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/22.jpg)
The integrating factor method (Sect. 2.1).
I Overview of differential equations.
I Linear Ordinary Differential Equations.I The integrating factor method.
I Constant coefficients.I The Initial Value Problem.I Variable coefficients.
![Page 23: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/23.jpg)
Linear Ordinary Differential Equations
Remark: Given a function y : R→ R, we use the notation
y ′(t) =dy
dt(t).
DefinitionGiven a function f : R2 → R, a first order ODE in the unknownfunction y : R→ R is the equation
y ′(t) = f (t, y(t)).
The first order ODE above is called linear iff there exist functionsa, b : R→ R such that f (t, y) = −a(t) y + b(t). That is, f islinear on its argument y , hence a first order linear ODE is given by
y ′(t) = −a(t) y(t) + b(t).
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Linear Ordinary Differential Equations
Remark: Given a function y : R→ R, we use the notation
y ′(t) =dy
dt(t).
DefinitionGiven a function f : R2 → R, a first order ODE in the unknownfunction y : R→ R is the equation
y ′(t) = f (t, y(t)).
The first order ODE above is called linear iff there exist functionsa, b : R→ R such that f (t, y) = −a(t) y + b(t). That is, f islinear on its argument y , hence a first order linear ODE is given by
y ′(t) = −a(t) y(t) + b(t).
![Page 25: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/25.jpg)
Linear Ordinary Differential Equations
Remark: Given a function y : R→ R, we use the notation
y ′(t) =dy
dt(t).
DefinitionGiven a function f : R2 → R, a first order ODE in the unknownfunction y : R→ R is the equation
y ′(t) = f (t, y(t)).
The first order ODE above is called linear iff there exist functionsa, b : R→ R such that f (t, y) = −a(t) y + b(t). That is, f islinear on its argument y , hence a first order linear ODE is given by
y ′(t) = −a(t) y(t) + b(t).
![Page 26: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/26.jpg)
Linear Ordinary Differential Equations
Example
A first order linear ODE is given by
y ′(t) = −2 y(t) + 3.
In this case function a(t) = −2 and b(t) = 3. Since these functiondo not depend on t, the equation above is called of constantcoefficients.
Example
A first order linear ODE is given by
y ′(t) = −2
ty(t) + 4t.
In this case function a(t) = −2/t and b(t) = 4t. Since thesefunctions depend on t, the equation above is called of variablecoefficients.
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Linear Ordinary Differential Equations
Example
A first order linear ODE is given by
y ′(t) = −2 y(t) + 3.
In this case function a(t) = −2 and b(t) = 3. Since these functiondo not depend on t, the equation above is called of constantcoefficients.
Example
A first order linear ODE is given by
y ′(t) = −2
ty(t) + 4t.
In this case function a(t) = −2/t and b(t) = 4t. Since thesefunctions depend on t, the equation above is called of variablecoefficients.
![Page 28: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/28.jpg)
Linear Ordinary Differential Equations
Example
A first order linear ODE is given by
y ′(t) = −2 y(t) + 3.
In this case function a(t) = −2 and b(t) = 3. Since these functiondo not depend on t, the equation above is called of constantcoefficients.
Example
A first order linear ODE is given by
y ′(t) = −2
ty(t) + 4t.
In this case function a(t) = −2/t and b(t) = 4t. Since thesefunctions depend on t, the equation above is called of variablecoefficients.
![Page 29: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/29.jpg)
Linear Ordinary Differential Equations
Example
A first order linear ODE is given by
y ′(t) = −2 y(t) + 3.
In this case function a(t) = −2 and b(t) = 3. Since these functiondo not depend on t, the equation above is called of constantcoefficients.
Example
A first order linear ODE is given by
y ′(t) = −2
ty(t) + 4t.
In this case function a(t) = −2/t and b(t) = 4t. Since thesefunctions depend on t, the equation above is called of variablecoefficients.
![Page 30: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/30.jpg)
The integrating factor method (Sect. 2.1).
I Overview of differential equations.
I Linear Ordinary Differential Equations.I The integrating factor method.
I Constant coefficients.I The Initial Value Problem.I Variable coefficients.
![Page 31: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/31.jpg)
The integrating factor method.
Remark: Solutions to first order linear ODE can be obtained usingthe integrating factor method.
Theorem (Constant coefficients)
Given constants a, b ∈ R with a 6= 0, the linear differential equation
y ′(t) = −a y(t) + b
has infinitely many solutions, one for each value of c ∈ R, given by
y(t) = c e−at +b
a.
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The integrating factor method.
Remark: Solutions to first order linear ODE can be obtained usingthe integrating factor method.
Theorem (Constant coefficients)
Given constants a, b ∈ R with a 6= 0, the linear differential equation
y ′(t) = −a y(t) + b
has infinitely many solutions, one for each value of c ∈ R, given by
y(t) = c e−at +b
a.
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The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
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The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
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The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above.
Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 36: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/36.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above.
Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 37: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/37.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′
⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 38: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/38.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 39: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/39.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y
⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 40: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/40.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′
⇔ µ′(t)
µ(t)= a.
![Page 41: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/41.jpg)
The integrating factor method.
Proof: Multiply the differential equation y ′(t) + a y(t) = b by anon-zero function µ, that is,
µ(t)(y ′ + ay
)= µ(t) b.
Key idea: The non-zero function µ is called an integrating factoriff holds
µ(y ′ + ay
)=
(µ y
)′.
Not every function µ satisfies the equation above. Let us find whatare the solutions µ of the equation above. Notice that
µ(y ′ + ay
)=
(µ y
)′ ⇔ µ y ′ + µay = µ′ y + µ y ′
ayµ = µ′ y ⇔ aµ = µ′ ⇔ µ′(t)
µ(t)= a.
![Page 42: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/42.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a.
Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 43: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/43.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a
⇔ ln(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 44: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/44.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 45: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/45.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0
⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 46: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/46.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 47: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/47.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .
For that function µ holds that µ(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 48: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/48.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′.
Therefore,multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 49: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/49.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ
⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 50: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/50.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat
⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 51: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/51.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 52: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/52.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c
⇔ y(t) = c e−at +b
a.
![Page 53: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/53.jpg)
The integrating factor method.
Proof: Recall:µ′(t)
µ(t)= a. Therefore,
[ln
(µ(t)
)]′= a ⇔ ln
(µ(t)
)= at + c0,
µ(t) = eat+c0 ⇔ µ(t) = eat ec0 .
Choosing the solution with c0 = 0 we obtain µ(t) = eat .For that function µ holds that µ
(y ′ + ay
)=
(µ y
)′. Therefore,
multiplying the ODE y ′ + ay = b by µ = eat we get
(µy)′ = bµ ⇔(eaty
)′= beat ⇔ eaty =
∫beat dt + c
y(t) eat =b
aeat + c ⇔ y(t) = c e−at +
b
a.
![Page 54: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/54.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 55: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/55.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 56: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/56.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 57: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/57.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 58: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/58.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 59: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/59.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 60: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/60.jpg)
The integrating factor method.
Example
Find all functions y solution of the ODE y ′ = 2y + 3.
Solution: The ODE is y ′ = −ay + b with a = −2 and b = 3.
The functions y(t) = ce−at +b
a, with c ∈ R, are solutions.
We conclude that the ODE hasinfinitely many solutions, given by
y(t) = c e2t − 3
2, c ∈ R.
Since we did one integration, it isreasonable that the solution contains aconstant of integration, c ∈ R.
0
y
t
−3/2c = 0
c < 0
c > 0
Verification: c e2t = y + (3/2), so 2c e2t = y ′, therefore weconclude that y satisfies the ODE y ′ = 2y + 3. C
![Page 61: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/61.jpg)
The integrating factor method (Sect. 2.1).
I Overview of differential equations.
I Linear Ordinary Differential Equations.I The integrating factor method.
I Constant coefficients.I The Initial Value Problem.I Variable coefficients.
![Page 62: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/62.jpg)
The Initial Value Problem.
DefinitionThe Initial Value Problem (IVP) for a linear ODE is the following:Given functions a, b : R→ R and constants t0, y0 ∈ R, find asolution y : R→ R of the problem
y ′ = a(t) y + b(t), y(t0) = y0.
Remark: The initial condition selects one solution of the ODE.
Theorem (Constant coefficients)
Given constants a, b, t0, y0 ∈ R, with a 6= 0, the initial valueproblem
y ′ = −ay + b, y(t0) = y0
has the unique solution
y(t) =(y0 −
b
a
)e−a(t−t0) +
b
a.
![Page 63: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/63.jpg)
The Initial Value Problem.
DefinitionThe Initial Value Problem (IVP) for a linear ODE is the following:Given functions a, b : R→ R and constants t0, y0 ∈ R, find asolution y : R→ R of the problem
y ′ = a(t) y + b(t), y(t0) = y0.
Remark: The initial condition selects one solution of the ODE.
Theorem (Constant coefficients)
Given constants a, b, t0, y0 ∈ R, with a 6= 0, the initial valueproblem
y ′ = −ay + b, y(t0) = y0
has the unique solution
y(t) =(y0 −
b
a
)e−a(t−t0) +
b
a.
![Page 64: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/64.jpg)
The Initial Value Problem.
DefinitionThe Initial Value Problem (IVP) for a linear ODE is the following:Given functions a, b : R→ R and constants t0, y0 ∈ R, find asolution y : R→ R of the problem
y ′ = a(t) y + b(t), y(t0) = y0.
Remark: The initial condition selects one solution of the ODE.
Theorem (Constant coefficients)
Given constants a, b, t0, y0 ∈ R, with a 6= 0, the initial valueproblem
y ′ = −ay + b, y(t0) = y0
has the unique solution
y(t) =(y0 −
b
a
)e−a(t−t0) +
b
a.
![Page 65: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/65.jpg)
The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0) = c − 3
2⇒ c =
5
2.
We conclude that y(t) =5
2e2t − 3
2. C
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The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0) = c − 3
2⇒ c =
5
2.
We conclude that y(t) =5
2e2t − 3
2. C
![Page 67: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/67.jpg)
The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0)
= c − 3
2⇒ c =
5
2.
We conclude that y(t) =5
2e2t − 3
2. C
![Page 68: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/68.jpg)
The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0) = c − 3
2
⇒ c =5
2.
We conclude that y(t) =5
2e2t − 3
2. C
![Page 69: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/69.jpg)
The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0) = c − 3
2⇒ c =
5
2.
We conclude that y(t) =5
2e2t − 3
2. C
![Page 70: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/70.jpg)
The Initial Value Problem.
Example
Find the solution to the initial value problem
y ′ = 2y + 3, y(0) = 1.
Solution: Every solution of the ODE above is given by
y(t) = c e2t − 3
2, c ∈ R.
The initial condition y(0) = 1 selects only one solution:
1 = y(0) = c − 3
2⇒ c =
5
2.
We conclude that y(t) =5
2e2t − 3
2. C
![Page 71: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/71.jpg)
The integrating factor method (Sect. 2.1).
I Overview of differential equations.
I Linear Ordinary Differential Equations.I The integrating factor method.
I Constant coefficients.I The Initial Value Problem.I Variable coefficients.
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The integrating factor method.
Theorem (Variable coefficients)
Given continuous functions a, b : R→ R and given constantst0, y0 ∈ R, the IVP
y ′ = −a(t)y + b(t) y(t0) = y0
has the unique solution
y(t) =1
µ(t)
[y0 +
∫ t
t0
µ(s)b(s)ds],
where the integrating factor function is given by
µ(t) = eA(t), A(t) =
∫ t
t0
a(s)ds.
Remark: See the proof in the Lecture Notes.
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The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
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The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 75: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/75.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 76: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/76.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t),
where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
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The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds
=
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 78: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/78.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds
= 2[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 79: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/79.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]
A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 80: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/80.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t)
= ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 81: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/81.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2)
⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 82: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/82.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 83: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/83.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: We first express the ODE as in the Theorem above,
y ′ = −2
ty + 4t.
Therefore, a(t) =2
tand b(t) = 4t, and also t0 = 1 and y0 = 2.
We first compute the integrating factor function µ = eA(t), where
A(t) =
∫ t
t0
a(s) ds =
∫ t
1
2
sds = 2
[ln(t)− ln(1)
]A(t) = 2 ln(t) = ln(t2) ⇒ eA(t) = t2.
We conclude that µ(t) = t2.
![Page 84: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/84.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2.
Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
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The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t)
⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 86: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/86.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 87: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/87.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3
⇔ t2y = t4 + c ⇔ y = t2 +c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 88: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/88.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c
⇔ y = t2 +c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 89: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/89.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 90: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/90.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1)
= 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 91: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/91.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c ,
that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 92: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/92.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 93: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/93.jpg)
The integrating factor method.
Example
Find the solution y to the IVP
t y ′ + 2y = 4t2, y(1) = 2.
Solution: The integrating factor is µ(t) = t2. Hence,
t2(y ′ +
2
ty)
= t2(4t) ⇔ t2 y ′ + 2t y = 4t3
(t2y
)′= 4t3 ⇔ t2y = t4 + c ⇔ y = t2 +
c
t2.
The initial condition implies 2 = y(1) = 1 + c , that is, c = 1.
We conclude that y(t) = t2 +1
t2. C
![Page 94: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/94.jpg)
Separable differential equations (Sect. 2.2).
I Separable ODE.
I Solutions to separable ODE.
I Explicit and implicit solutions.
I Homogeneous equations.
![Page 95: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/95.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 96: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/96.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 97: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/97.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)
⇔ f (t, y) =g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 98: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/98.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 99: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/99.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 100: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/100.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 101: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/101.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y)
and g(t) = −M(t).
![Page 102: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/102.jpg)
Separable ODE.
DefinitionGiven functions h, g : R→ R, a first order ODE on the unknownfunction y : R→ R is called separable iff the ODE has the form
h(y) y ′(t) = g(t).
Remark:A differential equation y ′(t) = f (t, y(t)) is separable iff
y ′ =g(t)
h(y)⇔ f (t, y) =
g(t)
h(y).
Notation:In lecture: t, y(t) and h(y) y ′(t) = g(t).
In textbook: x , y(x) and M(x) + N(y) y ′(x) = 0.
Therefore: h(y) = N(y) and g(t) = −M(t).
![Page 103: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/103.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable, since it isequivalent to(
1− y2)y ′(t) = t2 ⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 104: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/104.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable,
since it isequivalent to(
1− y2)y ′(t) = t2 ⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 105: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/105.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable, since it isequivalent to(
1− y2)y ′(t) = t2
⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 106: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/106.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable, since it isequivalent to(
1− y2)y ′(t) = t2 ⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 107: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/107.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable, since it isequivalent to(
1− y2)y ′(t) = t2 ⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.
Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 108: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/108.jpg)
Separable ODE.
Example
Determine whether the differential equation below is separable,
y ′(t) =t2
1− y2(t).
Solution: The differential equation is separable, since it isequivalent to(
1− y2)y ′(t) = t2 ⇒
{g(t) = t2,
h(y) = 1− y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = c t2, h(y) = c (1− y2), c ∈ R.
![Page 109: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/109.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable, since it isequivalent to
1
y2y ′(t) = − cos(2t) ⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 110: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/110.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable,
since it isequivalent to
1
y2y ′(t) = − cos(2t) ⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 111: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/111.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable, since it isequivalent to
1
y2y ′(t) = − cos(2t)
⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 112: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/112.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable, since it isequivalent to
1
y2y ′(t) = − cos(2t) ⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 113: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/113.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable, since it isequivalent to
1
y2y ′(t) = − cos(2t) ⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.
Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 114: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/114.jpg)
Separable ODE.
Example
Determine whether The differential equation below is separable,
y ′(t) + y2(t) cos(2t) = 0
Solution: The differential equation is separable, since it isequivalent to
1
y2y ′(t) = − cos(2t) ⇒
g(t) = − cos(2t),
h(y) =1
y2.
C
Remark: The functions g and h are not uniquely defined.Another choice here is:
g(t) = cos(2t), h(y) = − 1
y2.
![Page 115: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/115.jpg)
Separable ODE.
Remark: Not every first order ODE is separable.
Example
I The differential equation y ′(t) = ey(t) + cos(t) is notseparable.
I The linear differential equation y ′(t) = −2
ty(t) + 4t is not
separable.
I The linear differential equation y ′(t) = −a(t) y(t) + b(t),with b(t) non-constant, is not separable.
![Page 116: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/116.jpg)
Separable ODE.
Remark: Not every first order ODE is separable.
Example
I The differential equation y ′(t) = ey(t) + cos(t) is notseparable.
I The linear differential equation y ′(t) = −2
ty(t) + 4t is not
separable.
I The linear differential equation y ′(t) = −a(t) y(t) + b(t),with b(t) non-constant, is not separable.
![Page 117: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/117.jpg)
Separable ODE.
Remark: Not every first order ODE is separable.
Example
I The differential equation y ′(t) = ey(t) + cos(t) is notseparable.
I The linear differential equation y ′(t) = −2
ty(t) + 4t is not
separable.
I The linear differential equation y ′(t) = −a(t) y(t) + b(t),with b(t) non-constant, is not separable.
![Page 118: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/118.jpg)
Separable ODE.
Remark: Not every first order ODE is separable.
Example
I The differential equation y ′(t) = ey(t) + cos(t) is notseparable.
I The linear differential equation y ′(t) = −2
ty(t) + 4t is not
separable.
I The linear differential equation y ′(t) = −a(t) y(t) + b(t),with b(t) non-constant, is not separable.
![Page 119: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/119.jpg)
Separable differential equations (Sect. 2.2).
I Separable ODE.
I Solutions to separable ODE.
I Explicit and implicit solutions.
I Homogeneous equations.
![Page 120: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/120.jpg)
Solutions to separable ODE.
Theorem (Separable equations)
If the functions g , h : R→ R are continuous, with h 6= 0 and withprimitives G and H, respectively; that is,
G ′(t) = g(t), H ′(u) = h(u),
then, the separable ODE
h(y) y ′ = g(t)
has infinitely many solutions y : R→ R satisfying the algebraicequation
H(y(t)) = G (t) + c ,
where c ∈ R is arbitrary.
Remark: Given functions g , h, find their primitives G ,H.
![Page 121: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/121.jpg)
Solutions to separable ODE.
Theorem (Separable equations)
If the functions g , h : R→ R are continuous, with h 6= 0 and withprimitives G and H, respectively; that is,
G ′(t) = g(t), H ′(u) = h(u),
then, the separable ODE
h(y) y ′ = g(t)
has infinitely many solutions y : R→ R satisfying the algebraicequation
H(y(t)) = G (t) + c ,
where c ∈ R is arbitrary.
Remark: Given functions g , h, find their primitives G ,H.
![Page 122: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/122.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
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Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 124: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/124.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 125: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/125.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively,
are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 126: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/126.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 127: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/127.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 128: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/128.jpg)
Solutions to separable ODE.
Example
Find all solutions y : R→ R to the ODE y ′(t) =t2
1− y2(t).
Solution: The equation is equivalent to(1− y2
)y ′(t) = t2.
Therefore, the functions g , h are given by
g(t) = t2, h(u) = 1− u2.
Their primitive functions, G and H, respectively, are given by
g(t) = t2 ⇒ G (t) =t3
3,
h(u) = 1− u2 ⇒ H(u) = u − u3
3.
Then, the Theorem above implies that the solution y satisfies thealgebraic equation
y(t)− y3(t)
3=
t3
3+ c , c ∈ R. C
![Page 129: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/129.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement: Differentiate on bothsides with respect to t, that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3⇒ (1− y2) y ′ = t2.
![Page 130: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/130.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement: Differentiate on bothsides with respect to t, that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3⇒ (1− y2) y ′ = t2.
![Page 131: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/131.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement:
Differentiate on bothsides with respect to t, that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3⇒ (1− y2) y ′ = t2.
![Page 132: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/132.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement: Differentiate on bothsides with respect to t,
that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3⇒ (1− y2) y ′ = t2.
![Page 133: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/133.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement: Differentiate on bothsides with respect to t, that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3
⇒ (1− y2) y ′ = t2.
![Page 134: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/134.jpg)
Solutions to separable ODE.
Remarks:
I The equation y(t)− y3(t)
3=
t3
3+ c is algebraic in y , since
there is no y ′ in the equation.
I Every function y satisfying the algebraic equation
y(t)− y3(t)
3=
t3
3+ c ,
is a solution of the differential equation above.
I We now verify the previous statement: Differentiate on bothsides with respect to t, that is,
y ′(t)− 3(y2(t)
3
)y ′(t) = 3
t2
3⇒ (1− y2) y ′ = t2.
![Page 135: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/135.jpg)
Separable differential equations (Sect. 2.2).
I Separable ODE.
I Solutions to separable ODE.
I Explicit and implicit solutions.
I Homogeneous equations.
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Explicit and implicit solutions.
Remark:
The solution y(t)− y3(t)
3=
t3
3+ c is given in implicit form.
DefinitionAssume the notation in the Theorem above. The solution y of aseparable ODE is given in implicit form iff function y is specified by
H(y(t)
)= G (t) + c ,
The solution y of a separable ODE is given in explicit form ifffunction H is invertible and y is specified by
y(t) = H−1(G (t) + c
).
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Explicit and implicit solutions.
Remark:
The solution y(t)− y3(t)
3=
t3
3+ c is given in implicit form.
DefinitionAssume the notation in the Theorem above. The solution y of aseparable ODE is given in implicit form iff function y is specified by
H(y(t)
)= G (t) + c ,
The solution y of a separable ODE is given in explicit form ifffunction H is invertible and y is specified by
y(t) = H−1(G (t) + c
).
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Explicit and implicit solutions.
Remark:
The solution y(t)− y3(t)
3=
t3
3+ c is given in implicit form.
DefinitionAssume the notation in the Theorem above. The solution y of aseparable ODE is given in implicit form iff function y is specified by
H(y(t)
)= G (t) + c ,
The solution y of a separable ODE is given in explicit form ifffunction H is invertible and y is specified by
y(t) = H−1(G (t) + c
).
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Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
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Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable,
with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 141: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/141.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 142: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/142.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t)
⇔∫
y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 143: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/143.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 144: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/144.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt,
implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 145: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/145.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c
⇔ −1
u= −1
2sin(2t) + c .
![Page 146: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/146.jpg)
Explicit and implicit solutions.Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: The differential equation is separable, with
g(t) = − cos(2t), h(y) =1
y2.
The main idea in the proof of the Theorem above is this: integrateon both sides of the equation,
y ′(t)
y2(t)= − cos(2t) ⇔
∫y ′(t)
y2(t)dt = −
∫cos(2t) dt + c .
The substitution u = y(t), du = y ′(t) dt, implies that∫du
u2= −
∫cos(2t) dt + c ⇔ −1
u= −1
2sin(2t) + c .
![Page 147: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/147.jpg)
Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
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Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
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Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
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Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
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Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0)
=2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
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Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c,
so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
![Page 153: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/153.jpg)
Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
![Page 154: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/154.jpg)
Explicit and implicit solutions.
Example
Use the main idea in the proof of the Theorem above to find thesolution of the IVP
y ′(t) + y2(t) cos(2t) = 0, y(0) = 1.
Solution: Recall: −1
u= −1
2sin(2t) + c .
Substitute the unknown function y back in the equation above,
− 1
y(t)= −1
2sin(2t) + c . (Implicit form.)
y(t) =2
sin(2t)− 2c. (Explicit form.)
The initial condition implies that 1 = y(0) =2
0− 2c, so c = −1.
We conclude that y(t) =2
sin(2t) + 2. C
![Page 155: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/155.jpg)
Separable differential equations (Sect. 2.2).
I Separable ODE.
I Solutions to separable ODE.
I Explicit and implicit solutions.
I Homogeneous equations.
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Homogeneous equations.
DefinitionThe first order ODE y ′(t) = f
(t, y(t)
)is called homogeneous iff
for every numbers c , t, u ∈ R the function f satisfies
f (ct, cu) = f (t, u).
Remark:
I The function f is invariant under the change of scale of itsarguments.
I If f (t, u) has the property above, it must depend only on u/t.
I So, there exists F : R→ R such that f (t, u) = F(u
t
).
I Therefore, a first order ODE is homogeneous iff it has the form
y ′(t) = F(y(t)
t
).
![Page 157: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/157.jpg)
Homogeneous equations.
DefinitionThe first order ODE y ′(t) = f
(t, y(t)
)is called homogeneous iff
for every numbers c , t, u ∈ R the function f satisfies
f (ct, cu) = f (t, u).
Remark:
I The function f is invariant under the change of scale of itsarguments.
I If f (t, u) has the property above, it must depend only on u/t.
I So, there exists F : R→ R such that f (t, u) = F(u
t
).
I Therefore, a first order ODE is homogeneous iff it has the form
y ′(t) = F(y(t)
t
).
![Page 158: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/158.jpg)
Homogeneous equations.
DefinitionThe first order ODE y ′(t) = f
(t, y(t)
)is called homogeneous iff
for every numbers c , t, u ∈ R the function f satisfies
f (ct, cu) = f (t, u).
Remark:
I The function f is invariant under the change of scale of itsarguments.
I If f (t, u) has the property above, it must depend only on u/t.
I So, there exists F : R→ R such that f (t, u) = F(u
t
).
I Therefore, a first order ODE is homogeneous iff it has the form
y ′(t) = F(y(t)
t
).
![Page 159: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/159.jpg)
Homogeneous equations.
DefinitionThe first order ODE y ′(t) = f
(t, y(t)
)is called homogeneous iff
for every numbers c , t, u ∈ R the function f satisfies
f (ct, cu) = f (t, u).
Remark:
I The function f is invariant under the change of scale of itsarguments.
I If f (t, u) has the property above, it must depend only on u/t.
I So, there exists F : R→ R such that f (t, u) = F(u
t
).
I Therefore, a first order ODE is homogeneous iff it has the form
y ′(t) = F(y(t)
t
).
![Page 160: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/160.jpg)
Homogeneous equations.
DefinitionThe first order ODE y ′(t) = f
(t, y(t)
)is called homogeneous iff
for every numbers c , t, u ∈ R the function f satisfies
f (ct, cu) = f (t, u).
Remark:
I The function f is invariant under the change of scale of itsarguments.
I If f (t, u) has the property above, it must depend only on u/t.
I So, there exists F : R→ R such that f (t, u) = F(u
t
).
I Therefore, a first order ODE is homogeneous iff it has the form
y ′(t) = F(y(t)
t
).
![Page 161: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/161.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t. We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
) ⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 162: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/162.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t
⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t. We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
) ⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 163: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/163.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t. We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
) ⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 164: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/164.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t.
We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
) ⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 165: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/165.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t. We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
)
⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 166: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/166.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Rewrite the equation in the standard form
(t − y) y ′ = 2y − 3t − y2
t⇒ y ′ =
(2y − 3t − y2
t
)(t − y)
.
Divide numerator and denominator by t. We get,
y ′ =
(2y − 3t − y2
t
)(t − y)
(1
t
)(1
t
) ⇒ y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
![Page 167: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/167.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Recall: y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
We conclude that the ODE is homogeneous, because theright-hand side of the equation above depends only on y/t.
Indeed, in our case:
f (t, y) =2y − 3t − (y2/t)
t − y, F (x) =
2x − 3− x2
1− x,
and f (t, y) = F (y/t). C
![Page 168: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/168.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Recall: y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
We conclude that the ODE is homogeneous, because theright-hand side of the equation above depends only on y/t.
Indeed, in our case:
f (t, y) =2y − 3t − (y2/t)
t − y, F (x) =
2x − 3− x2
1− x,
and f (t, y) = F (y/t). C
![Page 169: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/169.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Recall: y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
We conclude that the ODE is homogeneous, because theright-hand side of the equation above depends only on y/t.
Indeed, in our case:
f (t, y) =2y − 3t − (y2/t)
t − y,
F (x) =2x − 3− x2
1− x,
and f (t, y) = F (y/t). C
![Page 170: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/170.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Recall: y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
We conclude that the ODE is homogeneous, because theright-hand side of the equation above depends only on y/t.
Indeed, in our case:
f (t, y) =2y − 3t − (y2/t)
t − y, F (x) =
2x − 3− x2
1− x,
and f (t, y) = F (y/t). C
![Page 171: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/171.jpg)
Homogeneous equations.
Example
Show that the equation below is homogeneous,
(t − y) y ′ − 2y + 3t +y2
t= 0.
Solution: Recall: y ′ =2(y
t
)− 3−
(y
t
)2
[1−
(y
t
)] .
We conclude that the ODE is homogeneous, because theright-hand side of the equation above depends only on y/t.
Indeed, in our case:
f (t, y) =2y − 3t − (y2/t)
t − y, F (x) =
2x − 3− x2
1− x,
and f (t, y) = F (y/t). C
![Page 172: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/172.jpg)
Homogeneous equations.
Example
Determine whether the equation below is homogeneous,
y ′ =t2
1− y3.
Solution:Divide numerator and denominator by t3, we obtain
y ′ =t2
(1− y3)
( 1
t3
)( 1
t3
) ⇒ y ′ =
(1
t
)( 1
t3
)−
(y
t
)3.
We conclude that the differential equation is not homogeneous. C
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Homogeneous equations.
Example
Determine whether the equation below is homogeneous,
y ′ =t2
1− y3.
Solution:Divide numerator and denominator by t3,
we obtain
y ′ =t2
(1− y3)
( 1
t3
)( 1
t3
) ⇒ y ′ =
(1
t
)( 1
t3
)−
(y
t
)3.
We conclude that the differential equation is not homogeneous. C
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Homogeneous equations.
Example
Determine whether the equation below is homogeneous,
y ′ =t2
1− y3.
Solution:Divide numerator and denominator by t3, we obtain
y ′ =t2
(1− y3)
( 1
t3
)( 1
t3
)
⇒ y ′ =
(1
t
)( 1
t3
)−
(y
t
)3.
We conclude that the differential equation is not homogeneous. C
![Page 175: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/175.jpg)
Homogeneous equations.
Example
Determine whether the equation below is homogeneous,
y ′ =t2
1− y3.
Solution:Divide numerator and denominator by t3, we obtain
y ′ =t2
(1− y3)
( 1
t3
)( 1
t3
) ⇒ y ′ =
(1
t
)( 1
t3
)−
(y
t
)3.
We conclude that the differential equation is not homogeneous. C
![Page 176: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/176.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
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Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 178: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/178.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F .
Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
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Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t.
This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 180: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/180.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t)
⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 181: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/181.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 182: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/182.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v)
⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 183: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/183.jpg)
Homogeneous equations.
TheoremIf the differential equation y ′(t) = f
(t, y(t)
)is homogeneous, then
the differential equation for the unknown v(t) =y(t)
tis separable.
Remark: Homogeneous equations can be transformed intoseparable equations.
Proof: If y ′ = f (t, y) is homogeneous, then it can be written asy ′ = F (y/t) for some function F . Introduce v = y/t. This means,
y(t) = t v(t) ⇒ y ′(t) = v(t) + t v ′(t).
Introducing all this into the ODE we get
v + t v ′ = F (v) ⇒ v ′ =
(F (v)− v
)t
.
This last equation is separable.
![Page 184: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/184.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 185: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/185.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
)
⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 186: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/186.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 187: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/187.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t,
soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 188: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/188.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′.
Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 189: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/189.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v
⇒ t v ′ =1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 190: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/190.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v
=1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 191: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/191.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 192: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/192.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: The equation is homogeneous, since
y ′ =t2 + 3y2
2ty
( 1
t2
)( 1
t2
) ⇒ y ′ =1 + 3
(y
t
)2
2(y
t
) .
Therefore, we introduce the change of unknown v = y/t, soy = t v and y ′ = v + t v ′. Hence
v + t v ′ =1 + 3v2
2v⇒ t v ′ =
1 + 3v2
2v− v =
1 + 3v2 − 2v2
2v
We obtain the separable equation v ′ =1
t
(1 + v2
2v
).
![Page 193: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/193.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
).
We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 194: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/194.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t
⇒∫
2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 195: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/195.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 196: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/196.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt,
so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 197: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/197.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0
⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 198: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/198.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0
⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 199: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/199.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 200: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/200.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 ,
so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 201: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/201.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t.
Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 202: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/202.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t
⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 203: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/203.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t
⇒ y(t) = ±t√
c1t − 1.
![Page 204: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/204.jpg)
Homogeneous equations.
Example
Find all solutions y of the ODE y ′ =t2 + 3y2
2ty.
Solution: Recall: v ′ =1
t
(1 + v2
2v
). We rewrite and integrate it,
2v
1 + v2v ′ =
1
t⇒
∫2v
1 + v2v ′ dt =
∫1
tdt + c0.
The substitution u = 1 + v2(t) implies du = 2v(t) v ′(t) dt, so∫du
u=
∫dt
t+ c0 ⇒ ln(u) = ln(t)+ c0 ⇒ u = e ln(t)+c0 .
But u = e ln(t)ec0 , so denoting c1 = ec0 , then u = c1t. Hence
1 + v2 = c1t ⇒ 1 +(y
t
)2= c1t ⇒ y(t) = ±t
√c1t − 1.
![Page 205: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/205.jpg)
Modeling with first order equations (Sect. 2.3).
I The mathematical modeling of natural processes.I Main example: Salt in a water tank.
I The experimental device.I The main equations.I Analysis of the mathematical model.I Predictions for particular situations.
![Page 206: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/206.jpg)
The mathematical modeling of natural processes.
Remarks:
I Physics describes natural processes with mathematicalconstructions, called physical theories.
I More often than not these physical theories containdifferential equations.
I Natural processes are described through solutions ofdifferential equations.
I Usually a physical theory, constructed to describe all knownnatural processes, predicts yet unknown natural processes.
I If the prediction is verified by an experiment or observation,one says that we have unveiled a secret from nature.
![Page 207: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/207.jpg)
The mathematical modeling of natural processes.
Remarks:
I Physics describes natural processes with mathematicalconstructions, called physical theories.
I More often than not these physical theories containdifferential equations.
I Natural processes are described through solutions ofdifferential equations.
I Usually a physical theory, constructed to describe all knownnatural processes, predicts yet unknown natural processes.
I If the prediction is verified by an experiment or observation,one says that we have unveiled a secret from nature.
![Page 208: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/208.jpg)
The mathematical modeling of natural processes.
Remarks:
I Physics describes natural processes with mathematicalconstructions, called physical theories.
I More often than not these physical theories containdifferential equations.
I Natural processes are described through solutions ofdifferential equations.
I Usually a physical theory, constructed to describe all knownnatural processes, predicts yet unknown natural processes.
I If the prediction is verified by an experiment or observation,one says that we have unveiled a secret from nature.
![Page 209: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/209.jpg)
The mathematical modeling of natural processes.
Remarks:
I Physics describes natural processes with mathematicalconstructions, called physical theories.
I More often than not these physical theories containdifferential equations.
I Natural processes are described through solutions ofdifferential equations.
I Usually a physical theory, constructed to describe all knownnatural processes, predicts yet unknown natural processes.
I If the prediction is verified by an experiment or observation,one says that we have unveiled a secret from nature.
![Page 210: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/210.jpg)
The mathematical modeling of natural processes.
Remarks:
I Physics describes natural processes with mathematicalconstructions, called physical theories.
I More often than not these physical theories containdifferential equations.
I Natural processes are described through solutions ofdifferential equations.
I Usually a physical theory, constructed to describe all knownnatural processes, predicts yet unknown natural processes.
I If the prediction is verified by an experiment or observation,one says that we have unveiled a secret from nature.
![Page 211: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/211.jpg)
Salt in a water tank.
Problem: Study the mass conservation law.
Particular situation: Salt concentration in water.
Main ideas of the test:
I Assuming the mass of salt and water is conserved, weconstruct a mathematical model for the salt concentration inwater.
I We study the predictions of this mathematical description.
I If the description agrees with the observation of the naturalprocess, then we conclude that the conservation of mass lawholds for salt in water.
![Page 212: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/212.jpg)
Salt in a water tank.
Problem: Study the mass conservation law.
Particular situation: Salt concentration in water.
Main ideas of the test:
I Assuming the mass of salt and water is conserved, weconstruct a mathematical model for the salt concentration inwater.
I We study the predictions of this mathematical description.
I If the description agrees with the observation of the naturalprocess, then we conclude that the conservation of mass lawholds for salt in water.
![Page 213: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/213.jpg)
Salt in a water tank.
Problem: Study the mass conservation law.
Particular situation: Salt concentration in water.
Main ideas of the test:
I Assuming the mass of salt and water is conserved, weconstruct a mathematical model for the salt concentration inwater.
I We study the predictions of this mathematical description.
I If the description agrees with the observation of the naturalprocess, then we conclude that the conservation of mass lawholds for salt in water.
![Page 214: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/214.jpg)
Salt in a water tank.
Problem: Study the mass conservation law.
Particular situation: Salt concentration in water.
Main ideas of the test:
I Assuming the mass of salt and water is conserved, weconstruct a mathematical model for the salt concentration inwater.
I We study the predictions of this mathematical description.
I If the description agrees with the observation of the naturalprocess, then we conclude that the conservation of mass lawholds for salt in water.
![Page 215: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/215.jpg)
Salt in a water tank.
Problem: Study the mass conservation law.
Particular situation: Salt concentration in water.
Main ideas of the test:
I Assuming the mass of salt and water is conserved, weconstruct a mathematical model for the salt concentration inwater.
I We study the predictions of this mathematical description.
I If the description agrees with the observation of the naturalprocess, then we conclude that the conservation of mass lawholds for salt in water.
![Page 216: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/216.jpg)
Modeling with first order equations (Sect. 2.3).
I The mathematical modeling of natural processes.I Main example: Salt in a water tank.
I The experimental device.I The main equations.I Analysis of the mathematical model.I Predictions for particular situations.
![Page 217: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/217.jpg)
The experimental device.
salt
r
r i i
o o
instantaneously mixed
V (t)
q (t)
q (t)
Q (t)
pipe
tank
water
![Page 218: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/218.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 219: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/219.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 220: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/220.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 221: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/221.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 222: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/222.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 223: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/223.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 224: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/224.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 225: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/225.jpg)
The experimental device.
Definitions:
I ri (t), ro(t): Rates in and out of water entering and leavingthe tank at the time t.
I qi (t), qo(t): Salt concentration of the water entering andleaving the tank at the time t.
I V (t): Water volume in the tank at the time t.
I Q(t): Salt mass in the tank at the time t.
Units:[ri (t)
]=
[ro(t)
]=
Volume
Time,
[qi (t)
]=
[qo(t)
]=
Mass
Volume.
[V (t)
]= Volume,
[Q(t)
]= Mass.
![Page 226: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/226.jpg)
Modeling with first order equations (Sect. 2.3).
I The mathematical modeling of natural processes.I Main example: Salt in a water tank.
I The experimental device.I The main equations.I Analysis of the mathematical model.I Predictions for particular situations.
![Page 227: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/227.jpg)
The main equations.
Remark: The mass conservation provides the main equations ofthe mathematical description for salt in water.
Main equations:
d
dtV (t) = ri (t)− ro(t), Volume conservation, (1)
d
dtQ(t) = ri (t) qi (t)− ro(t) qo(t), Mass conservation, (2)
qo(t) =Q(t)
V (t), Instantaneously mixed, (3)
ri , ro : Constants. (4)
![Page 228: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/228.jpg)
The main equations.
Remark: The mass conservation provides the main equations ofthe mathematical description for salt in water.
Main equations:
d
dtV (t) = ri (t)− ro(t), Volume conservation, (1)
d
dtQ(t) = ri (t) qi (t)− ro(t) qo(t), Mass conservation, (2)
qo(t) =Q(t)
V (t), Instantaneously mixed, (3)
ri , ro : Constants. (4)
![Page 229: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/229.jpg)
The main equations.
Remark: The mass conservation provides the main equations ofthe mathematical description for salt in water.
Main equations:
d
dtV (t) = ri (t)− ro(t), Volume conservation, (1)
d
dtQ(t) = ri (t) qi (t)− ro(t) qo(t), Mass conservation, (2)
qo(t) =Q(t)
V (t), Instantaneously mixed, (3)
ri , ro : Constants. (4)
![Page 230: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/230.jpg)
The main equations.
Remark: The mass conservation provides the main equations ofthe mathematical description for salt in water.
Main equations:
d
dtV (t) = ri (t)− ro(t), Volume conservation, (1)
d
dtQ(t) = ri (t) qi (t)− ro(t) qo(t), Mass conservation, (2)
qo(t) =Q(t)
V (t), Instantaneously mixed, (3)
ri , ro : Constants. (4)
![Page 231: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/231.jpg)
The main equations.
Remark: The mass conservation provides the main equations ofthe mathematical description for salt in water.
Main equations:
d
dtV (t) = ri (t)− ro(t), Volume conservation, (1)
d
dtQ(t) = ri (t) qi (t)− ro(t) qo(t), Mass conservation, (2)
qo(t) =Q(t)
V (t), Instantaneously mixed, (3)
ri , ro : Constants. (4)
![Page 232: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/232.jpg)
The main equations.
Remarks: [dV
dt
]=
Volume
Time=
[ri − ro
],
[dQ
dt
]=
Mass
Time=
[riqi − roqo
],
[riqi − roqo
]=
Volume
Time
Mass
Volume=
Mass
Time.
![Page 233: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/233.jpg)
Modeling with first order equations (Sect. 2.3).
I The mathematical modeling of natural processes.I Main example: Salt in a water tank.
I The experimental device.I The main equations.I Analysis of the mathematical model.I Predictions for particular situations.
![Page 234: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/234.jpg)
Analysis of the mathematical model.
Eqs. (4) and (1) imply
V (t) = (ri − ro) t + V0, (5)
where V (0) = V0 is the initial volume of water in the tank.
Eqs. (3) and (2) imply
d
dtQ(t) = ri qi (t)− ro
Q(t)
V (t). (6)
Eqs. (5) and (6) imply
d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t). (7)
![Page 235: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/235.jpg)
Analysis of the mathematical model.
Eqs. (4) and (1) imply
V (t) = (ri − ro) t + V0, (5)
where V (0) = V0 is the initial volume of water in the tank.
Eqs. (3) and (2) imply
d
dtQ(t) = ri qi (t)− ro
Q(t)
V (t). (6)
Eqs. (5) and (6) imply
d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t). (7)
![Page 236: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/236.jpg)
Analysis of the mathematical model.
Eqs. (4) and (1) imply
V (t) = (ri − ro) t + V0, (5)
where V (0) = V0 is the initial volume of water in the tank.
Eqs. (3) and (2) imply
d
dtQ(t) = ri qi (t)− ro
Q(t)
V (t). (6)
Eqs. (5) and (6) imply
d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t). (7)
![Page 237: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/237.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 238: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/238.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0,
and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 239: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/239.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 240: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/240.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 241: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/241.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q.
Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 242: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/242.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 243: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/243.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]
with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 244: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/244.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0,
where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 245: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/245.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t)
and A(t) =
∫ t
0a(s) ds.
![Page 246: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/246.jpg)
Analysis of the mathematical model.
Recall:d
dtQ(t) = ri qi (t)−
ro(ri − ro) t + V0
Q(t).
Notation: a(t) =ro
(ri − ro) t + V0, and b(t) = ri qi (t).
The main equation of the description is given by
Q ′(t) = −a(t) Q(t) + b(t).
Linear ODE for Q. Solution: Integrating factor method.
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b(s) ds
]with Q(0) = Q0, where µ(t) = eA(t) and A(t) =
∫ t
0a(s) ds.
![Page 247: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/247.jpg)
Modeling with first order equations (Sect. 2.3).
I The mathematical modeling of natural processes.I Main example: Salt in a water tank.
I The experimental device.I The main equations.I Analysis of the mathematical model.I Predictions for particular situations.
![Page 248: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/248.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 249: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/249.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).
In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 250: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/250.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0
⇒ a(t) =r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 251: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/251.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 252: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/252.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t)
⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 253: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/253.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 254: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/254.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 255: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/255.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Always holds Q ′(t) = −a(t) Q(t) + b(t).In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = rqi = b0.
We need to solve the IVP:
Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
![Page 256: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/256.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 257: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/257.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 258: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/258.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t,
µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 259: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/259.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t ,
Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 260: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/260.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 261: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/261.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)
⇒ Q(t) = e−a0t[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 262: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/262.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 263: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/263.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0.
Butb0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 264: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/264.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r
= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 265: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/265.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 266: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/266.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall the IVP: Q ′(t) = −a0 Q(t) + b0, Q(0) = Q0.
Integrating factor method:
A(t) = a0t, µ(t) = ea0t , Q(t) =1
µ(t)
[Q0 +
∫ t
0µ(s) b0 ds
].
∫ t
0µ(s) b0 ds =
b0
a0
(ea0t−1
)⇒ Q(t) = e−a0t
[Q0 +
b0
a0
(ea0t−1
)].
So: Q(t) =(Q0 −
b0
a0
)e−a0t +
b0
a0. But
b0
a0= rqi
V0
r= qiV0.
We conclude: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
![Page 267: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/267.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
Particular cases:
IQ0
V0> qi ;
IQ0
V0= qi , so Q(t) = Q0;
IQ0
V0< qi .
0
t
= q V0i
Q
0
Q0
Q
0
Q0
Q
Q
C
![Page 268: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/268.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
Particular cases:
IQ0
V0> qi ;
IQ0
V0= qi , so Q(t) = Q0;
IQ0
V0< qi .
0
t
= q V0i
Q
0
Q0
Q
0
Q0
Q
Q
C
![Page 269: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/269.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r , qi , Q0 and V0 are given, find Q(t).
Solution: Recall: Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0.
Particular cases:
IQ0
V0> qi ;
IQ0
V0= qi , so Q(t) = Q0;
IQ0
V0< qi .
0
t
= q V0i
Q
0
Q0
Q
0
Q0
Q
Q
C
![Page 270: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/270.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 271: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/271.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample.
Since Q(t) =(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 272: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/272.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0,
we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 273: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/273.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 274: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/274.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0
and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 275: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/275.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro ,
we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 276: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/276.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 277: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/277.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 .
Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 278: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/278.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1)
=Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 279: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/279.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0
⇒ e−rt1/V0 =1
100.
![Page 280: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/280.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: This problem is a particular case qi = 0 of the previousExample. Since Q(t) =
(Q0 − qiV0
)e−rt/V0 + qiV0, we get
Q(t) = Q0 e−rt/V0 .
Since V (t) = (ri − ro) t + V0 and ri = ro , we obtain V (t) = V0.
So q(t) = Q(t)/V (t) is given by q(t) =Q0
V0e−rt/V0 . Therefore,
1
100
Q0
V0= q(t1) =
Q0
V0e−rt1/V0 ⇒ e−rt1/V0 =
1
100.
![Page 281: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/281.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100.
Then,
− r
V0t1 = ln
( 1
100
)= − ln(100) ⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 282: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/282.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100. Then,
− r
V0t1 = ln
( 1
100
)
= − ln(100) ⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 283: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/283.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100. Then,
− r
V0t1 = ln
( 1
100
)= − ln(100)
⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 284: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/284.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100. Then,
− r
V0t1 = ln
( 1
100
)= − ln(100) ⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 285: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/285.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100. Then,
− r
V0t1 = ln
( 1
100
)= − ln(100) ⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 286: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/286.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r and qi are constants.If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter,find t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value.
Solution: Recall: e−rt1/V0 =1
100. Then,
− r
V0t1 = ln
( 1
100
)= − ln(100) ⇒ r
V0t1 = ln(100).
We conclude that t1 =V0
rln(100).
In this case: t1 = 100 ln(100). C
![Page 287: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/287.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 288: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/288.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t).
In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 289: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/289.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0
⇒ a(t) =r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 290: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/290.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 291: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/291.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t)
⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 292: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/292.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 293: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/293.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 294: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/294.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds,
µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 295: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/295.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.
![Page 296: The integrating factor method (Sect. 2.1)....The integrating factor method (Sect. 2.1). I Overview of differential equations. I Linear Ordinary Differential Equations. I The integrating](https://reader030.fdocuments.in/reader030/viewer/2022040621/5f346255461c387d410b4bce/html5/thumbnails/296.jpg)
Predictions for particular situations.
Example
Assume that ri = ro = r are constants. If r = 5x106 gal/year,qi (t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, find Q(t).
Solution: Recall: Q ′(t) = −a(t) Q(t) + b(t). In this case:
a(t) =ro
(ri − ro) t + V0⇒ a(t) =
r
V0= a0,
b(t) = ri qi (t) ⇒ b(t) = r[2 + sin(2t)
].
We need to solve the IVP: Q ′(t) = −a0 Q(t) + b(t), Q(0) = 0.
Q(t) =1
µ(t)
∫ t
0µ(s) b(s) ds, µ(t) = ea0t ,
We conclude: Q(t) = re−rt/V0
∫ t
0ers/V0
[2 + sin(2s)
]ds.