Solve the differential equation by ** method of undetermined coefficient** .
$ y''-5y'+6y=e^{t} \cos 2t+e^{2t} (3t+4) \sin t \ $
Answer:
The auxiliary equation is
$ m^2-5m+6=0 \\ \Rightarrow m=2,3 $
The complementary function is
$ C.F.=c_1 e^{2t}+c_2 e^{3t} \ $ , where $ \ c_1, c_2 \ $ are arbitrary constants
But I can not assume how to construct the particular integral $ \ P.I. \ $
I think the particular integral $ \ P.I.=Ae^{t} \cos 2t +B t^2 e^{2t} (3t+4) \sin t \ $ ,
because $ \ e^{2t} \ $ appears in the complimentary function .
But I am not sure.
Help me out
As Mario pointed out you can split the equation into two equation $$ \begin{cases} y''-5y'+6y=e^{t} \cos 2t \\ y''-5y'+6y=e^{2t} (3t+4) \sin t \end{cases} $$ Substitute for the first $y=ze^t$ and for the second $y=ue^{2t}$
Then simplify both equations $$ \begin{cases} z''-z'+2z=\cos 2t \\ u''-u'= (3t+4) \sin t \end{cases} $$ Solve both equations to get the particular solution
Edit for youmath
you get $$ \begin{cases} z=z_h+z_p \\ u=u_h+u_p \end{cases} \implies \begin{cases} y=z_he^t+z_pe^t \\ y=u_he^{2t}+u_pe^{2t} \end{cases} \implies y=e^tz_h+z_pe^t +u_he^{2t}+u_pe^{2t}$$
The homegeneous part you already have it for y but now you have the particular part too.... $$y_p=z_pe^t +u_pe^{2t}$$