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OpenStudy (anonymous):
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)?
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OpenStudy (anonymous):
Find two particular solutions, then add the up. One for y''+4y = xe^x and one for y''+4y = xsin(2x)
OpenStudy (anonymous):
The first one should be of the form \[ y_{p1}= (Ax + B ) e^x \]
OpenStudy (anonymous):
Ok, @Loser66 help him.
OpenStudy (anonymous):
Excellent
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