Linear second order non-homogeneous ODE question

[Malby]

Determine the general solution to the ODE:

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

I know the solution will be of the form y = y_h + y_p. 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:

y_p = ((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

 

[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

 

[Malby]

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

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

Cheers

 

[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 y_h since they're already being used as the constants of integration for y_h.

 

[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

 

[Malby]

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

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

Thanks very much!