General solution to non-homogenous differential y''-2y'=e^(2x) + x^2 - 1

[BigHairyMommaBear]

I can find the general solution, y_c, to y''-2y'=0 which I believe to be y_c=A+Be^2x.
I then go on to try and find the particular solution, y_p, to y''-2y'=e^2x using y_p=pe^2x which gives y'_p=2pe^2x and y''_p=4pe^2x. Substituting these values into the original equation I get 4pe^2x-2(2pe^2x)=e^2x which then give 4pe^2x-4pe^2x=e^2x which is clearly not true.
I'm not sure whether I'm making a mistake here or whether I have to use a different method in order to find the general solution. Any help would be appreciated.

 

[Deleted member 4993]

I can find the general solution, y_c, to y''-2y'=0 which I believe to be y_c=A+Be^2x.
I then go on to try and find the particular solution, y_p, to y''-2y'=e^2x using y_p=pe^2x which gives y'_p=2pe^2x and y''_p=4pe^2x. Substituting these values into the original equation I get 4pe^2x-2(2pe^2x)=e^2x which then give 4pe^2x-4pe^2x=e^2x which is clearly not true.
I'm not sure whether I'm making a mistake here or whether I have to use a different method in order to find the general solution. Any help would be appreciated.

since e^(2x) is a homogeneous solution, the particular solution should be x*e^(2x).