CHAPTER 1 - FIRST ORDER DIFFERENTIAL EQUATIONS · 2018-10-08 · FIRST ORDER DIFFERENTIAL EQUATIONS...

BASIC CONCEPTS SEPARATION OF VARIABLES EQUATIONS WITH HOMOGENEOUS COEFFICIENTS EXACT DIFFERENTIAL EQUATIONS LINEAR DIFFERENTIAL EQUATIONS Integrating Factors Found By Inspection The General Procedure for Determining the Integrating Factor Coefficients Linear in x and y

CHAPTER 1FIRST ORDER DIFFERENTIAL EQUATIONS

Differential Equations

DIFEQUA DLSU-Manila

I Definition: A differential equation is an equation thatcontains a function and one or more of its derivatives. Ifthe function has only one independent variable, then it isan ordinary differential equation. Otherwise, it is apartial differential equation.

I The following are examples of differential equations:

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

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(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

(a)∂2u∂x2 +

∂2u∂y2 = 0

(b) (x2 + y2)dx − 2xydy = 0

(c)d3xdy3

+ xdxdy− 4xy = 0

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

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Order and DegreeI Definition: The order of a differential equation is the order

of the highest ordered derivative that appears in the givenequation. The degree of a differential equation is thedegree of the highest ordered derivative treated as avariable.

I Examples:

(a)∂2u∂x2 +

∂2u∂y2 = 0 is of order 2 and degree 1

(b) (x2 + y2)dx − 2xydy = 0 is of order 1 and degree 1

d3xdy3

+ xdxdy− 4xy = 0 is of order 3 and degree 2

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

is of order 2 and degree 3

DIFEQUA DLSU-Manila

I Examples:

(a)∂2u∂x2 +

d3xdy3

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

I Examples:

(a)∂2u∂x2 +

d3xdy3

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

I Examples:

(a)∂2u∂x2 +

d3xdy3

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

I Examples:

(a)∂2u∂x2 +

d3xdy3

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

I Examples:

(a)∂2u∂x2 +

d3xdy3

(d)∂u∂t

= h2(∂2u∂x2 +

∂2u∂y2

DIFEQUA DLSU-Manila

Solution of a Differential EquationI Definition: A solution of a differential equation is a

function defined explicitly or implicitly by an equation thatsatisfies the given equation. The general solutionrepresents all the possible solutions of the given equation,while a particular solution is any one of the possiblesolutions of a given differential equation.

I Examples:

(a)dydx

= y , y = Cex where C is an arbitrary constant

(b)dydx

= 3ex , y = 3ex + C where C is an arbitrary constant

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

(e)d2ydt2 + k2y = 0, y = sin kt , where k is a constant

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I Examples:

(a)dydx

(b)dydx

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

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I Examples:

(a)dydx

(b)dydx

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

DIFEQUA DLSU-Manila

I Examples:

(a)dydx

(b)dydx

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

DIFEQUA DLSU-Manila

I Examples:

(a)dydx

(b)dydx

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

DIFEQUA DLSU-Manila

I Examples:

(a)dydx

(b)dydx

(c) y (3) − 3y ′ + 2y = 0, y = e−2x

(d)dydx

=−(x + 1)

y − 3, (x + 1)2 + (y − 3)2 = c2, where c is an

arbitrary constant.

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Existence and Uniqueness TheoremI The existence of a particular solution satisfying initial

conditions of the form y(x0) = y0 is guaranteed by thefollowing theorem:

I Existence and Uniqueness Theorem: Consider a first orderequation of the form

= f (x , y)

and let T be the rectangular region described by

T = { (x , y) ∈ R2 | |x−x0| ≤ a, |y−y0| ≤ b, a,b are positive constants }.If f and fy are continuous in T , then there exists a positivenumber h and a function y = y(x) such that(a) y = y(x) is a solution of the given equation satisfying

y(x0) = y0; and(b) y = y(x) is unique on the interval |x − x0| ≤ h.

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Existence and Uniqueness TheoremI The existence of a particular solution satisfying initial

conditions of the form y(x0) = y0 is guaranteed by thefollowing theorem:

I Existence and Uniqueness Theorem: Consider a first orderequation of the form

= f (x , y)

and let T be the rectangular region described by

T = { (x , y) ∈ R2 | |x−x0| ≤ a, |y−y0| ≤ b, a,b are positive constants }.If f and fy are continuous in T , then there exists a positivenumber h and a function y = y(x) such that(a) y = y(x) is a solution of the given equation satisfying

y(x0) = y0; and(b) y = y(x) is unique on the interval |x − x0| ≤ h.

DIFEQUA DLSU-Manila

Exercises

(1) Verify that the given function is a solution of the givendifferential equation.(a) y (3) − 3y ′ + 2y = 0, y = e−2x

(b)d2ydt2 + k2y = 0, y = sin kt , where k is a constant

(2) Use antiderivatives to obtain a general or a particularsolution to each of the following equations:

(a)dydx

= x3 + 2x

(b)dydx

= 4 cos 2x

(c)dydx

= 3ex , y = 6 x = 0

(d)dydx

= 4y , y = 3 whenx = 0

DIFEQUA DLSU-Manila

Separable Differential EquationI A first order ordinary differential equation can generally be

expressed in the form

Mdx + Ndy = 0

where M and N may be functions of two variables x and y .For this reason, we call this the general form of afirst-order ordinary differential equation.

I If the differential equation can be manipulated algebraicallyto transform it into an equivalent form

A(x)dx + B(y)dy = 0

where A(x) is a function of x alone and B(y) is a functionof y alone, then we say that the variables x and y areseparable.

DIFEQUA DLSU-Manila

Separable Differential EquationI A first order ordinary differential equation can generally be

expressed in the form

Mdx + Ndy = 0

where M and N may be functions of two variables x and y .For this reason, we call this the general form of afirst-order ordinary differential equation.

I If the differential equation can be manipulated algebraicallyto transform it into an equivalent form

A(x)dx + B(y)dy = 0

where A(x) is a function of x alone and B(y) is a functionof y alone, then we say that the variables x and y areseparable.

DIFEQUA DLSU-Manila

Examples

I Show that the variables in the differential equation(1− x)y ′ = y2 are separable.

I Find the general solution of the equation xyy ′ = 1 + y2, ify = 3 when x = 2.

I Find the particular solution of the equation y ′ = −2xysatisfying y(0) = 1.

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Examples

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Examples

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Exercises

(1) y ′ = y sec x(2) sin x sin y dx + cos x cos y dy = 0(3) y ln x ln y dx + dy = 0(4) (xy + x) dx = (x2y2 + x2 + y2 + 1) dy(5) dx = t(1 + t2) sec2 x dt(6) (e2x + 4)y ′ = y(7) x2 dx + y(x − 1) dy = 0(8) 2xyy ′ = 1 + y2, y = 3 when x = 2(9) 2y dx = 3x dy , y = −1 when x = 2

(10) y ′ = x exp(y − x2), y = 0 when x = 0

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Homogeneous Functions

I Definition: Let λ be a parameter. A function f (x , y) is saidto be homogeneous of degree k , where k is a real number,if f satisfies the condition

f (λx , λy) = λk f (x , y)

I Example: Show that the function f (x , y) = 4x2 − 3xy + y2

is homogeneous and determine the degree ofhomogeneity.

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Homogeneous Functions

I Definition: Let λ be a parameter. A function f (x , y) is saidto be homogeneous of degree k , where k is a real number,if f satisfies the condition

f (λx , λy) = λk f (x , y)

I Example: Show that the function f (x , y) = 4x2 − 3xy + y2

is homogeneous and determine the degree ofhomogeneity.

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Two Theorems

I Theorem: If M(x , y) and N(x , y) are both homogeneousfunctions of the same degree, then the function

g(x , y) =M(x , y)N(x , y)

is homogeneous of degree zero.

I Theorem: If f (x , y) is homogeneous of degree zero, then fis a function of y/x (or x/y ) alone.

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Two Theorems

I Theorem: If M(x , y) and N(x , y) are both homogeneousfunctions of the same degree, then the function

g(x , y) =M(x , y)N(x , y)

is homogeneous of degree zero.

I Theorem: If f (x , y) is homogeneous of degree zero, then fis a function of y/x (or x/y ) alone.

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Equation With Homogeneous Coefficients

I Definition: A first order differential equation of the formM dx +N dy = 0 is said to have homogeneous coefficientsif M and N are homogeneous functions of the samedegree.

I Example: Show that the equation3(3x2 + y2) dx − 2xy dy = 0 has homogeneouscoefficients.

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Equation With Homogeneous Coefficients

I Definition: A first order differential equation of the formM dx +N dy = 0 is said to have homogeneous coefficientsif M and N are homogeneous functions of the samedegree.

I Example: Show that the equation3(3x2 + y2) dx − 2xy dy = 0 has homogeneouscoefficients.

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Solution of an Equation With HomogeneousCoefficients

Given: Mdx + Ndy = 0(a) Transform the equation into the equivalent form

g(y/x) + dydx = 0, where g = M/N.

(b) Let y = vx and use this to convert the second equation intothe form x dv + (v + g(v)) dx = 0 where the variables areseparable.

(c) Solve using the method of separation of variables.(d) Use v = y/x to obtain a solution in terms of x and y .(e) Remark: You may also use the substitution

x = vy ,dx = v dy + y dv .

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Examples

(a) Solve the equation 3(3x2 + y2) dx − 2xy dy = 0(b) Show that the equation

y dx = (x +√

y2 − x2) dy

has homogeneous coefficients and find the general solutionusing the method discussed in this section.

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Examples

y dx = (x +√

y2 − x2) dy

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Examples

y dx = (x +√

y2 − x2) dy

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Exercises

Find the general solution of each of the following equations:(1) 3xy dx + (x2 + y2) dy = 0(2) [x csc(y/x)− y ] dx + x dy = 0(3) xy dx − (x + 2y)2 dy = 0(4) (3x2 − 2xy + 3y2) dx = 4xy dy(5) [x − y tan−1(y/x)] dx + x tan−1(y/x) dy = 0

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Exercises

Find the particular solution indicated for each of the followingequations:(1) (x − y) dx + (3x + y) dy = 0 if y = −2 when x = 3

(2) (y −√

x2 + y2)dx − x dy = 0 if y = 1 when x = 0

(3) xy dx + 2(x2 + 2y2) dy = 0 if y = 1 when x = 0(4) (3x2 − 2y2)y ′ = 2xy if y = −1 when x = 0

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Exact Differential EquationI Definition: A differential equation of the form

M(x , y) dx + N(x , y) dy = 0 is said to be exact if thereexists a differentiable function F (x , y) whose totaldifferential is

dF = M(x , y) dx + N(x , y) dy = 0

I Example: Let F (x , y) = x3y − 2x2y2 + 3xy . Then

dF =∂F∂x

dx +∂F∂y

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy

I This shows that the first order differential equation

(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0

is exact, since its left-hand side is exactly dF .DIFEQUA DLSU-Manila

dF = M(x , y) dx + N(x , y) dy = 0

dF =∂F∂x

dx +∂F∂y

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy

(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0

dF = M(x , y) dx + N(x , y) dy = 0

dF =∂F∂x

dx +∂F∂y

= (3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy

(3x2y − 4xy2 + 3y) dx + (x3 − 4x2y + 3x) dy = 0

Remark

I Remark: If M(x , y) dx + N(x , y) dy = 0 is exact, thenthere exists a function F such thatdF = M(x , y) dx + N(x , y) dy . This means that thegeneral solution is of the form F (x , y) = C, where C is anarbitrary constant.

I Example: Consider the simple differential equationy dx + x dy = 0.

I The left-hand side of this equation is easily seen to be thetotal differential of the function F (x , y) = xy , so that ourdifferential equation takes the form dF = 0.

I Integration then yields the general solution xy = c.

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Remark

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Remark

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Remark

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Necessary and Sufficient Condition for Exactness

I If M, N,∂M∂y

and∂N∂x

are all continuous functions of x and

y , then the differential equationM(x , y) dx + N(x , y) dy = 0 is exact if and only if

∂M∂y

=∂N∂x

I Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact.

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Necessary and Sufficient Condition for Exactness

I If M, N,∂M∂y

and∂N∂x

are all continuous functions of x and

y , then the differential equationM(x , y) dx + N(x , y) dy = 0 is exact if and only if

∂M∂y

=∂N∂x

I Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact.

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Solution of an Exact Differential Equation

I The general solution of an exact equationM dx + N dy = 0 is of the form F (x , y) = C.

I Since∂F∂x

= M,∂F∂y

, the function F can be obtained by integration and therelation

∂M∂y

=∂N∂x

I Example: Show that the equation(6x + y2) dx + y(2x − 3y) dy = 0 is exact, and find thegeneral solution.

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I Since∂F∂x

= M,∂F∂y

∂M∂y

=∂N∂x

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I Since∂F∂x

= M,∂F∂y

∂M∂y

=∂N∂x

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Additional Examples

(1) Find a particular solution for the differential equation

2xy dx + (x2 + y2) dy = 0

if y = 1 when x = −1.(2) Find a function M(x , y) so that the differential equation

M(x , y) dx +

(xexy + 2xy +

)dy = 0

is exact.

DIFEQUA DLSU-Manila

Additional Examples

(1) Find a particular solution for the differential equation

2xy dx + (x2 + y2) dy = 0

if y = 1 when x = −1.(2) Find a function M(x , y) so that the differential equation

M(x , y) dx +

(xexy + 2xy +

)dy = 0

is exact.

DIFEQUA DLSU-Manila

Exercises

(1) Solve the following differential equations:(a) (2xy − 3x2) dx + (x2 + y) dy = 0(b) (y2 − 2xy + 6x) dx − (x2 − 2xy + 2) dy = 0(c) (1 + y2 + xy2) dx + (x2y + y + 2xy) dy = 0(d) (xy2 + x − 2y + 3) dx + x2y dy = 2(x + y) dy , y = 1 when

x = 1(e) (3y(x2 − 1) dx + (x3 + 8y − 3x) dy = 0, y = 1 when x = 0

(2) Find a function N(x , y) such that the equation(y1/2x−1/2 +

xx2 + y

)dx + N(x , y) dy = 0

will be exact.

DIFEQUA DLSU-Manila

Linear Differential EquationI Definition: A linear differential equation of order one is an

equation which is expressible in the form

A(x)dydx

+ B(x)y = C(x)

where A, B and C are all functions of the single variable x .Equations of this type may also be written in the form

+ P(x)y = Q(x).

This equation is linear with respect to y .I Examples:

(a) The equation y ′ + 3x2y = x2 is linear with P(x) = 3x2 andQ(x) = x2.

(b) Show that the equation 2(y − 4x2) dx + x dy = 0 is linearwith respect to y .

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A(x)dydx

+ B(x)y = C(x)

+ P(x)y = Q(x).

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A(x)dydx

+ B(x)y = C(x)

+ P(x)y = Q(x).

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A(x)dydx

+ B(x)y = C(x)

+ P(x)y = Q(x).

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Equations Linear With Respect to xI Remark: An equation which is not linear in y may be linear

with respect to x . In this case, the equation may be writtenin the form

+ P(y)x = Q(y)

I In general, an equation containing y dy can not be linear iny , and an equation containing x dx is not linear in x .

I Example: The equation (x + 4y2)dy + 2y dx = 0 can bewritten in the equivalent form

)x = −2y

which is linear with respect to x .

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+ P(y)x = Q(y)

)x = −2y

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+ P(y)x = Q(y)

)x = −2y

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The Integrating Factor

I Definition: A linear equation can be made exact bymultiplying it with an appropriate function v . This functionis called an integrating factor for the linear differentialequation (with respect to the variable y ) of the formv(x) > 0.

I When multiplied to the above equation, the integratingfactor v(x) produces an exact equation of the form

+ P(x)y)

= Q(x)v(x)

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The Integrating Factor

I Definition: A linear equation can be made exact bymultiplying it with an appropriate function v . This functionis called an integrating factor for the linear differentialequation (with respect to the variable y ) of the formv(x) > 0.

I When multiplied to the above equation, the integratingfactor v(x) produces an exact equation of the form

+ P(x)y)

= Q(x)v(x)

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Finding the Integrating FactorI

v dy + vPy dx = vQ dx(vPy − vQ) dx + v dy = 0

I Since the equation is exact, with M = vPy − vQ andN = v , we should have

∂M∂y

= vP =∂N∂x

= v ′(x)

from which we get P dx =dvv

where the variables havebeen separated.

I Antidifferentiation gives the integrating factor

v = exp( ∫

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∂M∂y

= vP =∂N∂x

= v ′(x)

v = exp( ∫

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∂M∂y

= vP =∂N∂x

= v ′(x)

v = exp( ∫

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Procedure for Solving a Linear Differential Equation

I Write the equation in the formdydx

+ P(x)y = Q(x).

I Solve for the integrating factor v = exp( ∫

I Multiply the equation in (1) by the integrating factor v toobtain an exact differential equation.

I Solve the resulting differential equation.I Example: Show that the equation y ′ = x − 2y is linear, and

find the general solution.

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+ P(x)y = Q(x).

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+ P(x)y = Q(x).

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+ P(x)y = Q(x).

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+ P(x)y = Q(x).

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+ P(x)y = Q(x).

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RemarkI In the preceding example, the left hand side of the

equation e2x dydx

+ 2e2xy = xe2x is of the formddx

(e2xy),so that we get the equation

(e2xy) = xe2x

which is exact because the solution can be obtainedimmediately by integration.

I More generally, if the linear equationdydx

+ Py = Q is

multiplied by the integrating factor v = exp(∫

obtain vdydx

+ vPy = Qv where the left hand side is ddx (vy)

and the right hand side is a function of x . This is exact andthe solution is obtained by integrating both sides.DIFEQUA DLSU-Manila

equation e2x dydx

(e2xy) = xe2x

+ Py = Q is

obtain vdydx

equation e2x dydx

(e2xy) = xe2x

+ Py = Q is

obtain vdydx

Equations Linear in x

I In general, an equation which contains an expression ofthe form yk dy is not linear with respect to y . It may,however, be linear with respect to x if it can be expressedin the form

+ P(y)x = Q(y).

I In this case, the integrating factor is a function of y

obtained from the expression v = exp( ∫

P(y) dy)

I Show that the equation 2y(y2 − x) dy = dx is linear withrespect to x , and find its general solution.

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+ P(y)x = Q(y).

P(y) dy)

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+ P(y)x = Q(y).

P(y) dy)

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Exercises

i Find the general solution of each of the following equations:(a) (x5 + 3y) dx − x dy = 0

(b)dxdy

+1− 3y

yx = 3

(c) 2y dx = (x2 − 1)(dx − dy)(d) (y + 1) dx + (4x − y) dy = 0(e) (y − cos2 x) dx + cos x dy = 0(f) (y − x + xy cot x) dx + x dy = 0

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Some Common Exact Differentials

I Some exact differentials frequently occur within first orderordinary differential equations. Some of these exactdifferentials are the following:(a) d(xy) = x dy + y dx

(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

x dy − y dxx2 + y2

I These exact differentials are easily recognized because oftheir distinct forms.

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(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

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(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

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(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

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(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

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(b) d(

y dx − x dyy2

(c) d(y

x dy − y dxx2

(d) d(

tan−1 yx

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Solution by Inspection

I We consider differential equations in which some termscan be grouped together such that one or more of somecommon differentials can be identified.

I Using an appropriate change of variables and/orintegrating factor, a simple differential equation where thevariables are separable can be obtained.

I This method of solving a differential equation is calledsolution by inspection.

I Examples:(a) y(y3 − x) dx + x(y3 + x) dy = 0(b) y(y2 + 1) dx + x(y2 − 1)dy = 0(c) y(x3 − y5) dx − x(x3 + y5) dy = 0

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Exercises

Solve the following differential equations by inspection.(a) y(2xy + 1) dx − x dy = 0(b) (x3y3 + 1) dx + x4y2 dy = 0(c) y(x4 − y2) dx + x(x4 + y2) dy = 0(d) y(x2y2 − 1) dx + x(x2y2 + 1) dy = 0(e) y(2x + y2) dx + x(y2 − x) dy = 0(f) y(x2 + y2 − 1) dx + x(x2 + y2 + 1) dy = 0

(g) y(x3exy − y) dx + x(y + x3exy ) dy = 0

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The General Procedure for Determining theIntegrating Factor

I Let u be an integrating factor for an equation of the formM dx + N dy = 0.

I Then the equation uM dx + uN dy = 0 is exact. Thismeans there exists a function F (x , y) such that

∂F∂x

= uM,∂F∂y

I This means that u must satisfy the equation

u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x

or, equivalently, the equation

u(∂M∂y− ∂N∂x

∂u∂x−M

∂u∂y

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∂F∂x

= uM,∂F∂y

u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x

∂u∂x−M

∂u∂y

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∂F∂x

= uM,∂F∂y

u∂M∂y

+ M∂u∂y

= u∂N∂x

+ N∂u∂x

∂u∂x−M

∂u∂y

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Special Cases

I If1N

(∂M∂y− ∂N∂x

)= f (x) is a function of x alone, then the

integrating factor is of the form

u = exp∫

f (x) dx

I If1M

(∂M∂y− ∂N∂x

)= g(y) is a function of y alone, then

the integrating factor is of the form

u = exp∫−g(y) dy

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Special Cases

I If1N

(∂M∂y− ∂N∂x

)= f (x) is a function of x alone, then the

integrating factor is of the form

u = exp∫

f (x) dx

I If1M

(∂M∂y− ∂N∂x

)= g(y) is a function of y alone, then

the integrating factor is of the form

u = exp∫−g(y) dy

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Examples

Solve the following differential equations:(a) 2y(x2 − y + x) dx + (x2 − 2y) dy = 0(b) y2 dx + (3xy + y2 − 1) dy = 0(c) (4xy + y2) dx + (2y − 2x2) dy = 0

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Exercises

Find the appropriate integrating factors to solve the followingequations:(a) (x2 + y2 + 1) dx + (x2 − 2xy) dy = 0(b) (4xy + y2) dx + (2y − 2x2) dy = 0(c) (y2 + 2xy − 2y) dx − (2x + 2y) dy = 0(d) (2x2y − xy2 + y) dx + (x − y) dy = 0(e) y(2x − y + 1) dx + x(3x − 4y + 3) dy = 0(f) 2(2y2 + 5xy − 2y + 4) dx + (2x2 + 2xy − x) dy = 0

(g) (2y2 + 3xy − 2y + 6x) dx + x(x + 2y − 1) dy = 0

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Coefficients Linear in x and y

I We consider equations of the form(a1x + b1y + c1) dx + (a2x + b2y + c2) dy = 0.

I We will introduce a change of variables which will reducethe above equation to a homogeneous equation.

I We consider the associated linesa1x + b1y + c1 = 0a2x + b2y + c2 = 0

I We consider the following cases:Case 1: The lines are intersecting.Case 2: The lines are parallel.

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Case 1: The lines intersect at a point (h, k)

I Consider the change of variables x = u + h, y = v + k ,so that dx = du and dy = dv .

I Our given equation transforms into

(au + b1v) du + (a2u + b2v) dv = 0

which has homogeneous coefficients.I Example: Solve the equation

(x − 4y − 3) dx − (x − 6y − 5) dy = 0.

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(au + b1v) du + (a2u + b2v) dv = 0

(x − 4y − 3) dx − (x − 6y − 5) dy = 0.

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(au + b1v) du + (a2u + b2v) dv = 0

(x − 4y − 3) dx − (x − 6y − 5) dy = 0.

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Case 2: The associated lines are parallel

I In this case, the coefficients of the variables in theequations of the two lines are proportional, so there existsa constant k such that a2x + b2y = k(a1x + b1y),

I Letting u = a1x + b1y will transform the equation into anequivalent equation where the variables are separable.

I Example: Solve the equation(6x − 3y + 2) dx − (2x − y − 1) dy = 0.

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Exercises

Solve the following equations:(a) (y − 2) dx − (x − y − 1) dy = 0(b) (x − 4y − 9) dx + (4x + y − 2) dy = 0(c) (x + y − 1) dx + (2x + 2y + 1) dy = 0(d) (x − 3y + 2) dx + 3(x + 3y − 4) dy = 0

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CHAPTER 1 - FIRST ORDER DIFFERENTIAL EQUATIONS · 2018-10-08 · FIRST ORDER DIFFERENTIAL EQUATIONS...

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