I am trying to find the particular solution to the $y^{(4)} -2y'' +y = xe^x $ and currently am misunderstanding what to do.
My steps:
the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $ which 0 at 1: $p(1) = 0 $
so now i know that the solution will be something like:
$y_p = (Ax^3 + Bx^2)e^x$ where the parentheses show that it is a linear operator on the last coefficient.
so far I think that my method is to differentiate this four times and collect up the terms of each $y_p$ according to the original equation and see what A and B are equal to.
I assume this is okay to do with linear operators?
$y_p' = (Ax^3 + Bx^2)e^x + (3Ax^2 + 2Bx)e^x$
$y_p'' = (Ax^3 +(6A + B)x^2 + (4B + 6A)x + 2B)e^x$
$y_p''' = (Ax^3 +(9A +B)x^2 + (18A +6B)x + (6A + 6B))e^x$
$y_p'''' = (Ax^3 +(12A +B)x^2 + (36A +8B)x + (24A + 12B))e^x$
which then gives me an equation of $A-2A +A = 1$ which is wrong because this is the equation for the first term of the particular solution. Can anyone please help me take it from here?