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yaro99
Having trouble with the particular solution: y''+4y'+8y=2cos(2t)+sin(2t)
I tried \[y_p=A \cos 2t+B \sin 2t\] and plugged in its derivatives, but I ended up with A=2/13 and B=3/13, but in the answer they are A=0 and B=1/4
Oops I the 2 in 2t when differentiating. That's probably it, gonna try it now.
I neglected* the 2 in 2t