Originally Posted by

**tttcomrader** Find the particular solution to $\displaystyle y^{(4)}+5y''+4y = \sin x + \cos 2x $

First, I found that the general solution is $\displaystyle y_h=$$\displaystyle C_1 \cos (2x) + C_2 \sin (2x) + C_3 \cos (-2x) +$$\displaystyle C_4 \sin (-2x) + C_5 \cos (x) + C_6 \sin (x) + C_7 \cos (-x) + C_8 \sin (-x) $

So the list of terms that I obtain here is: $\displaystyle \{ \cos (2x), \sin (2x), \cos (-2x) , \sin (-2x) , \cos (x) , \sin (x) , \cos (-x) , \sin (-x) \} $

Now, for my guess of the particular solution, I have list of terms $\displaystyle \{ \sin x , \cos (2x) \} $

Since I have repeats, I would take the derivative, so I have new list $\displaystyle \{ \sin x , \cos x , \cos (2x) , \sin (2x) \} $

So my particular solution is: $\displaystyle y_p = A \sin x + B \cos x + C \cos (2x) + D \sin (2x) $

I know I'm missing something, I think I'm suppose to multiply x by one of the list, but I just can't remember what the method was, please help. Thanks.