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)?

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)?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Ok, @Loser66 help him.

anonymous · Answer

Excellent