particular solution to $y^{(4)} -2y'' +y = xe^x $ using undetermined coefficients

Question

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?

$y_p' = (Ax^3 + Bx^2)e^x + (3Ax^2 + 2Bx)e^x$

$y_p'' = (Ax^3 +(6A + B)x^2 + (4B + 6A)x + 2B)e^x$

can I collect terms from different linear operators like i did with $y_p''$

and I assume that to find the particular solution I must find $y_p''''$ as it is in the original equation

Answer 1

Yes, you must find $y_p''$ and $y_p''''$ and collect terms, then solve for the coefficients of $e^x$ and $x e^x$ on left and right sides to be equal.

Answer 2

This is correct as the comments point out.

A good way to differentiate the expression $(Ax^3+Bx^2)e^x$ is to render it as just $u(x)e^x$ and use rge product rule:

$y_p=ue^x$

$y_p'=(u+u')e^x$

$y_p''=((u+u')+(u'+u''))e^x=(u+2u'+u'')e^x$

It becomes apparent that the coefficients on the derivatives of $u$ match the rows of Pascal's Triangle.

$y_p'''=(u+3u'+3u''+3u''')e^x$

$y_p''''=(u+4u'+6u''+4u'''+u'''')e^x$

Plug these into your derivative combination $y''''-2y''+y$ and put back $u=Ax^3+Bx^2$, then match up the result with the desired quantity $xe^x$.