# Linear second order non-homogeneous ODE question

1. Jul 31, 2011

### Malby

Determine the general solution to the ODE:

y'' + 2y' = 1 + xe-2x

I know the solution will be of the form y = yh + yp. The homogeneous solution is y = c1 + c2e-2x.

For the particular solution, I have been using the method of undetermined coefficients. c3e-2x won't work as it is not linearly independent of the homogeneous solution. So I guess c3xe-2x.

So y' = c3e-2x(c3 - 2c3x)

And y'' = 4c3e-2x(x - 1)

Following on from this I end up with a particular solution:

yp = ((1 + xe-2x)/(-2e-2x))xe-2x

However wolfram disagrees with me. My question is, is the method of undetermined coefficients ok to use for this ODE? Or should I be using the method with the Wronskian (it's name escapes me at the moment)

Cheers

2. Jul 31, 2011

### Ray Vickson

If you put y(x) = C1(x) + C2(x)*exp(-2x) in the DE, you eventually arrive at a first-order DE in v = dC1/dx + exp(-2x)*dC2/dx.

RGV

3. Jul 31, 2011

### Malby

I'm not sure I understand this. Are you able to explain it a little more?

Cheers

4. Aug 1, 2011

### Bohrok

I'd have to dig out my book or notes to make sure, but I think your particular solution will also need an e-2x term, so it should be of the form Ae-2x + Bxe-2x. I wouldn't use c's for yh since they're already being used as the constants of integration for yh.

5. Aug 1, 2011

### ehild

For the method of variation of the parameters, read http://en.wikipedia.org/wiki/Variation_of_parameters.

This problem can be solved for y' as the y term is missing. Denote z=y' and solve the first-order equation z'+2z=1+xe-2x, then integrate z=y' to get y.

ehild

6. Aug 1, 2011

### Malby

Aha! Of course. It's all so simple once you know what to do... :rofl:

Thanks very much!