hello,

i have problem solving this one

$\displaystyle y'' - 2y' = e^{-x} (2\cos {x} + 3 \sin {x} ) $

okay i don't have any problem with homogeneous part

$\displaystyle r^2-2r=0 $

$\displaystyle r(r-2)=0 $

$\displaystyle r_1=0 \; ; \; r_2=2$

$\displaystyle y_h = C_1 e^{0x} + C_2 e^{2x} $

$\displaystyle y_h = C_1 + C_2 e^{2x} $

i have problem with particular part now... (never encountered with similar till now) and really don't know where to go from here ...

i know that if instead of what i have here is :

$\displaystyle y'' -2y' = \cos {x} + e^x $

particular part would be

$\displaystyle y_p = A\cos{x} + B \sin {x} + Ce^x $

but with that up there i'm just stuck ...

any help is very much appreciated