I read that when the right side $f(x) = e^x$, a suggested form of $y_{ps}$ is $Ce^x$, and for $f(x)$ is linear in $x$, a form of $y_{ps}$ is $Cx + D$. If $f(x)$ is the product of the two mentioned, then the form of $y_{ps}$ is also the product of their particular solutions. Source.
I'm working on the following example, and having trouble finding the particular solution: $$ y'' - y' -2y = (2x-1)e^x $$
If I understood correctly, I could treat this problem as two parts: $$ y'' - y' -2y = (2x-1) $$ and $$ y'' - y' -2y = e^x $$ Then find the particular solutions for both of them, then multiply the particular solutions, would that be correct?
You can just use $y_p=(Ax+B)e^x$
Expand it you'll get $Axe^x+Be^x$ Then find the first and second derivatives and plug them into the original funtion