What's the answer to this 2nd order linear differential equation

d^2y/dx^2 + 4dy/dx+4y= 8e^(2x)

I got y(x)= yh(x) + yp(x)= (A+Bx)e^(-2x)- 100/27 x^2 e^(2x)

but I dont think that's right, especially the last bit...

y'' + 4y' + 4y = 8e^(2x)

Well it's non-homogeneous so the solution is in y = yc + yp form where yc is the homogeneous case and yp is the accounting for the other possible solutions. (I'm using different variables but they're the same)

So let's do the homogeneous part first... (yc)
y'' + 4y' + 4y = 0
(I use r to substitute for the y'',y')
r^2 + 4r + 4 = 0
(r + 2)(r + 2) = 0
r = -2
yc = c1e^(rx) + c2xe^(rx)
yc = c1e^(-2x)+c2xe^(-2x)
{where c1 and c2 are just constants)

Now for yp, the Longer part:
so we choose yp = Ae^(2x)
We need to find y''p, y'p using yp because we need to substitute them in.
yp = Ae^(2x)
y'p = 2Ae^(2x)
y''p = 4Ae^(2x)

Substitute:
y''p + 4y'p + 4yp = 8e^(2x)
4Ae^(2x) + 4(2Ae^(2x))+4Ae^(2x) = 8e^(2x)
Solve for A:
4Ae^(2x) + 8Ae^(2x) +4 Ae^(2x) = 8e^(2x)
16Ae^(2x) = 8 e^(2x)
A = 8/16
A = 1/2

So... yp = Ae^(2x)
Plug in A: yp = (1/2)e^(2x)

Now that we have both yc and yp, we can find the general solution y(x) = yc + yp:
y = c1e^(-2x)+c2xe^(-2x) + (1/2)e^(2x)

Here's my answer, along with the answer that Wolfram Alpha gives (which is consistent with my answer).

http://img535.imageshack.us/img535/6962/…
http://img194.imageshack.us/img194/4504/…

Wolfram Alpha Solution:
http://www.wolframalpha.com/input/?i=y%2…


Hope this helps.

The last bit find in the form
C e^(2x)

It will work because eigenvalues are indeed -2, but the exponent on the RHS is +2
Look more closely