- #1

Malby

- 16

- 0

y'' + 2y' = 1 + xe

^{-2x}

I know the solution will be of the form y = y

_{h}+ y

_{p}. The homogeneous solution is y = c

_{1}+ c

_{2}e

^{-2x}.

For the particular solution, I have been using the method of undetermined coefficients. c

_{3}e

^{-2x}won't work as it is not linearly independent of the homogeneous solution. So I guess c

_{3}xe

^{-2x}.

So y' = c

_{3}e

^{-2x}(c

_{3}- 2c

_{3}x)

And y'' = 4c

_{3}e

^{-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