Lecture 6: Integrating factors

Amath 351 Winter 2019

Instructor: Chay Paterson

January 23, 2019

Topics

1. Integrating factors

1 Integrating factors (skill 5)

Consider a general first-order ODE is of the form

a1(x)y′+a0(x)y+g(x) = 0.

Dividing by a1(x) and rearranging some terms gives an equivalent ODE of the form

y′+p(x)y=q(x)

for some functions p(x) and q(x). We will learn how to solve any equation of this form, and therefore any

first order ODE! We have a few cases to consider.

•Suppose q(x) = 0 for all x. Then we are back in the separable case!

•Next, suppose p(x) = 0 for all x. This equation is still separable and is even easier to solve.

dx=q(x) =⇒y=Zq(x) dx+c.

•Let us look at a specific example, where neither of these functions is zero. First, recall the Fundamental

Theorem of Calculus. If I have any differentiable function F(z), we have RF′(z)dz =F(z)+c. Second,

I am going to work on an example and apply some tricks which will seem odd, but bear with me.

Consider the ODE y′+y= 5. Then, we have

ex(y′+y) = 5ex

=⇒exy′+exy= 5ex

=⇒(exy)′= 5ex(using the product rule)

=⇒Z(exy)′dx=Z5exdx

=⇒exy= 5ex+c

=⇒y=e−x5 + ce−x

=⇒y= 5 + ce−x.

What was the primary trick we used? Multiplying by ex.

Why did that help? Because it allowed use to write the left-hand side as a single derivative (using the

product rule). And because it’s easy to integrate a single derivative.

AMATH 351 Winter Quarter 2019 Lecture 6

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