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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