Question

What is the particular solution to the equation: y''+4y = xe^x + xsin(2x). I am confused in supposing the particular solution for both terms on RHS. For xe^x the annihilating factor is (D-1)^2 i guess. but what for the xsin(2x)?

Answer 1

Find two particular solutions, then add the up.
One for
y''+4y = xe^x
and one for
y''+4y = xsin(2x)

Answer 2

The first one should be of the form
\[
y_{p1}= (Ax + B ) e^x
\]

Answer 3

Ok, @Loser66 help him.

Answer 4

Excellent