# Integrating factor for a 2nd order homogeneous linear ODE

1. Mar 18, 2013

### danny_91

1. The problem statement, all variables and given/known data
Consider the general linear homogeneous second order equation:
P(x)y'' + Q(x)y' + R(x)y = 0 (1)

We seek an integrating factor μ(x) such that, upon multiplying Eq. (1) by μ(x), we can write the resulting equation in the form
[μ(x)P(x)y']' + μ(x)R(x)y = 0 (2)

(a) By equating coefficients of y' in Eqs. (1) and (2), show that μ must be a solution of:
Pμ' = (Q - P')μ (3)

2. Relevant equations
I guess the only relevant equation is the product rule of differentiation:
[a(x)b(x)]' = a'(x)b(x) + a(x)b'(x)

3. The attempt at a solution
Here is what I have done so far: By setting (1) = (2)

P(x)y'' + Q(x)y' + R(x)y = [μ(x)P(x)y']' + μ(x)R(x)y
Py'' + Qy' + Ry = μPy'' + μP'y' + μ'Py' + μRy
Py'' + Qy' + Ry = μPy'' + (μP' + μ'P)y' + μRy

Equating coefficients of y':

Q = μP' + μ'P
So: Pμ' = Q - P'μ, which is not the form that the book stated, Pμ' = (Q - P')μ

So am I missing something in the derivation, or is it that the book's statement is wrong?
By the way, this is Problem 11 in Section 11.1 of the book called Elementary Differential Equations and Boundary Value Problems by Boyce & DiPrima (9th Edition).

Part (b) of the problem is just solving the differential equation to find μ(x), and I'm pretty sure that I can do that part. Anyway, I would really appreciate any input.

2. Mar 18, 2013

### BruceW

I don't think you can equate the coefficients of the differentials of y, because of equation (1) which gives an explicit relationship between them. So you can't just treat them as separate things as you normally would.

Your method is good, apart from that. You should try again. There are 3 things y,y',y'' that you don't want in your final formula, so try to get rid of one of them by substituting it from one equation into the other.

3. Mar 20, 2013

### danny_91

Thank you so much, BruceW! :)

So I went back to this step: Py'' + Qy' + Ry = μPy'' + (μP' + μ'P)y' + μRy, and then grouping like terms of y'', y', and y. And then I use the relationship from Eq. (1) then I'm left with just y'. So, yes, I got the answer as the book stated.

Thanks so much for pointing out that simply equating the coefficients of y' is not the right approach!

4. Mar 22, 2013

### BruceW

no problem :) in a way, this question is pretty tricky, since it is really pretty lucky that the y'' and y happen to both fall out. (It is not obvious from the beginning what is the right way to go). But you've got it, so all's good!