Help with annihilators

[Romeo]

So I have to find the general solution to a differential equation. The equation is y''+ y = 6e^x.
I know that y''+ y = (D^2 +1)y and I know the annihilator for 6e^x is (D-1). So I multiplied them together and I get D^3 - D^2 + D - 1 which I have reduced to ((D-1)^2)(D+1).
My y(x)= c1e^-x + c2e^x + axe^x.
I have my y_p = axe^x
y_p' = ae^x(x+1)
y_p'' = ae^x(x+2)
and when I try to calculate an answer for a, I always end up with an x value for some reason.
Can anyone help me?

 

[HallsofIvy]

You multiplied together \(\displaystyle (D^2+ 1)(D- 1)\) and got something that factored to \(\displaystyle (D- 1)^2(D+ 1)\)?

That is impossible! \(\displaystyle (D^2+ 1)(D- 1)\) is the factored form.

What is the general solution to y''+ y= 0? To get \(\displaystyle 6e^x\), try something like \(\displaystyle y= Ae^x\). What must A be?