First, the characteristic solution. The homogeneous ODE
[tex]y''-3y'+2y=0[/tex]
has characteristic equation
[tex]r^2-3r+2=(r-2)(r-1)=0[/tex]
which has roots at [tex]r=2[/tex] and [tex]r=1[/tex], which means the characteristic solution is
[tex]y_c=C_1e^{2x}+C_2e^x[/tex]
To find the particular solution, the method of undetermined coefficients will probably be the easiest to use. Suppose the particular solution and its derivative are
[tex]y_p=ae^{-x}+b\cos3x+c\sin3x[/tex]
[tex]{y_p}'=-ae^{-x}-3b\sin3x+3c\cos3x[/tex]
[tex]{y_p}''=ae^{-x}-9b\cos3x-9c\sin3x[/tex]
Substituting into the ODE gives
[tex]\left(ae^{-x}-9b\cos3x-9c\sin3x\right)-3\left(-ae^{-x}-3b\sin3x+3c\cos3x\right)+2\left(ae^{-x}+b\cos3x+c\sin3x\right)=3e^{-x}-10\cos3x[/tex]
[tex]6ae^{-x}+(-9b-9c+2b)\cos3x+(-9c+9b+2c)\sin3x=3e^{-x}-10\cos3x[/tex]
[tex]6ae^{-x}+(-7b-9c)\cos3x+(-7c+9b)\sin3x=3e^{-x}-10\cos3x[/tex]
It follows that
[tex]\begin{cases}6a=3\\-7b-9c=-10\\-7c+9b=0\end{cases}\implies a=\dfrac12,b=\dfrac7{13},c=\dfrac9{13}[/tex]
So the solution is
[tex]y=y_c+y_p[/tex]
[tex]y=C_1e^{2x}+C_2e^x+\dfrac12e^{-x}+\dfrac7{13}\cos3x+\dfrac9{13}\sin3x[/tex]
