Post on 28-Jan-2021
September 4 Math 2306 sec. 51 Fall 2019
Section 4: First Order Equations: Linear
A first order linear equation has the form
a1(x)dydx
+ a0(x)y = g(x).
If g(x) = 0 the equation is called homogeneous. Otherwise it is callednonhomogeneous.
Provided a1(x) 6= 0 on the interval I of definition of a solution, we canwrite the standard form of the equation
dydx
+ P(x)y = f (x).
We’ll be interested in equations (and intervals I) for which P and f arecontinuous on I.
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Solutions (the General Solution)
dydx
+ P(x)y = f (x).
It turns out the solution will always have a basic form of y = yc + ypwhere
I yc is called the complementary solution and would solve theequation
dydx
+ P(x)y = 0
(called the associated homogeneous equation), andI yp is called the particular solution, and is heavily influenced by
the function f (x).
The cool thing is that our solution method will get both parts in oneprocess—we won’t get this benefit with higher order equations!
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Motivating Example
x2dydx
+2xy = ex
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Derivation of Solution via Integrating Factor
Solve the equation in standard form
dydx
+ P(x)y = f (x)
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General Solution of First Order Linear ODE
I Put the equation in standard form y ′ + P(x)y = f (x), and correctlyidentify the function P(x).
I Obtain the integrating factor µ(x) = exp(∫
P(x)dx).
I Multiply both sides of the equation (in standard form) by theintegrating factor µ. The left hand side will always collapse intothe derivative of a product
ddx
[µ(x)y ] = µ(x)f (x).
I Integrate both sides, and solve for y .
y(x) =1
µ(x)
∫µ(x)f (x)dx = e−
∫P(x) dx
(∫e∫
P(x) dx f (x)dx + C)
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Solve the ODE
dydx
+y = 3xe−x
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Solve the IVP
xdydx−y = 2x2, x > 0 y(1) = 5
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VerifyJust for giggles, lets verify that our solution y = 2x2 + 3x really doessolve the differential equation we started with
xdydx− y = 2x2.
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Steady and Transient States
For some linear equations, the term yc decays as x (or t) grows. Forexample
dydx
+ y = 3xe−x has solution y =32
x2e−x + Ce−x .
Here, yp =32
x2e−x and yc = Ce−x .
Such a decaying complementary solution is called a transient state.
The corresponding particular solution is called a steady state.
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Bernoulli Equations
Suppose P(x) and f (x) are continuous on some interval (a,b) and n isa real number different from 0 or 1 (not necessarily an integer). Anequation of the form
dydx
+ P(x)y = f (x)yn
is called a Bernoulli equation.
Observation: This equation has the flavor of a linear ODE, but sincen 6= 0,1 it is necessarily nonlinear. So our previous approach involvingan integrating factor does not apply directly. Fortunately, we can use achange of variables to obtain a related linear equation.
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Solving the Bernoulli Equation
dydx
+ P(x)y = f (x)yn
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Example
Solve the initial value problem y ′ − y = −e2xy3, subject to y(0) = 1.
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Exact Equations
We considered first order equations of the form
M(x , y)dx + N(x , y)dy = 0. (1)
The left side is called a differential form. We will assume here that Mand N are continuous on some (shared) region in the plane.
Definition: The equation (1) is called an exact equation on somerectangle R if there exists a function F (x , y) such that
∂F∂x
= M(x , y) and∂F∂y
= N(x , y)
for every (x , y) in R.
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Exact Equation Solution
If M(x , y)dx + N(x , y)dy = 0 happens to be exact, then it isequivalent to
∂F∂x
dx +∂F∂y
dy = 0
This implies that the function F is constant on R and solutions to the
DE are given by the relation
F (x , y) = C
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Recognizing Exactness
There is a theorem from calculus that ensures that if a function F hasfirst partials on a domain, and if those partials are continuous, then thesecond mixed partials are equal. That is,
∂2F∂y∂x
=∂2F∂x∂y
.
If it is true that∂F∂x
= M and∂F∂y
= N
this provides a condition for exactness, namely
∂M∂y
=∂N∂x
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Exact Equations
Theorem: Let M and N be continuous on some rectangle R in theplane. Then the equation
M(x , y)dx + N(x , y)dy = 0
is exact if and only if∂M∂y
=∂N∂x
.
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ExampleShow that the equation is exact and obtain a family of solutions.
(2xy − sec2 x)dx + (x2 + 2y)dy = 0
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Special Integrating Factors
Suppose that the equation M dx + N dy = 0 is not exact. Clearly ourapproach to exact equations would be fruitless as there is no suchfunction F to find. It may still be possible to solve the equation if wecan find a way to morph it into an exact equation. As an example,consider the DE
(2y − 6x)dx + (3x − 4x2y−1)dy = 0
Note that this equation is NOT exact. In particular
∂M∂y
= 2 6= 3− 8xy−1 = ∂N∂x
.
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Special Integrating FactorsBut note what happens when we multiply our equation by the functionµ(x , y) = xy2.
xy2(2y − 6x)dx + xy2(3x − 4x2y−1)dy = 0, =⇒
(2xy3 − 6x2y2)dx + (3x2y2 − 4x3y)dy = 0
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Special Integrating Factors
The function µ is called a special integrating factor. Finding one(assuming one even exists) may require ingenuity and likely a bit ofluck. However, there are certain cases we can look for and perhapsuse them to solve the occasional equation. A useful method is to lookfor µ of a certain form (usually µ = xnym for some powers n and m).We will restrict ourselves to two possible cases:
There is an integrating faction µ = µ(x) depending only on x , or thereis an integrating factor µ = µ(y) depending only on y .
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Special Integrating Factor µ = µ(x)
Suppose thatM dx + N dy = 0
is NOT exact, but that
µM dx + µN dy = 0
IS exact where µ = µ(x) does not depend on y . Then
∂(µ(x)M)∂y
=∂(µ(x)N)
∂x.
Let’s use the product rule in the right side.
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Special Integrating Factor µ = µ(x)
∂(µ(x)M)∂y
=∂(µ(x)N)
∂x.
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Special Integrating Factor
M dx + N dy = 0 (2)
Theorem: If (∂M/∂y − ∂N/∂x)/N is continuous and depends only onx , then
µ = exp
(∫ ∂M∂y −
∂N∂x
Ndx
)is an special integrating factor for (2). If (∂N/∂x − ∂M/∂y)/M is
continuous and depends only on y , then
µ = exp
(∫ ∂N∂x −
∂M∂y
Mdy
)
is an special integrating factor for (2).
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Example
Solve the equation 2xy dx + (y2 − 3x2)dy = 0.
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