Find the general solution of: $y^"+3y^{'}+2y=\sin(x)$ (I'm using the Annihilator Method) The characteristic eqn corresponding to LHS is $m^2+3m+2=0$. The solutions are: $m = -2$ and $m = -1$. So $y_c=Ae^{-2x}+Be^{-x}$. 

The Annihilator of $sin x = D^2+1 $. Which makes $y_p=Ccosx +Dsinx$. Where $A, B, C, D$ are constants. 

To find the constants $C$, and $D$. I found out $y_p^{'} =-Csinx+Dcosx$ 
    and $y_p^{''} =-Ccosx-Dsinx$. I put these values in the original eqn ($y^"+3y^{'}+2y=sinx$). I'm not getting the right values. C should be -3/10 and D should be 1/10. 

Any clues as to where I'm going wrong?

Thanks.
$$y^"+3y^{'}+2y=\sin(x)$$
Using annihilator 's method
$$(D+1)(D+2)(D^2+1)y=0$$
$$
\begin{align}
y_h=&amp; \, c_1e^{-x}+c_2e^{-2x} \\
y_p=&amp; \, C\sin(x)+D\cos(x)
\end{align}
$$
$$(-C\sin(x)-D\cos(x)+3(C\cos(x)-D\sin(x))+2(C\sin(x)+D\cos(x))=\sin(x)$$
$$
\begin{cases}
-D+3C+2D=0 \\
-C-3D+2C=1 
\end{cases}
\implies 
\begin{cases}
D=-3C \\
C-3D=1 
\end{cases}
\implies 
\begin{cases}
D=-3/10\\
C=1/10 
\end{cases}
$$
$$\boxed{y_p=\frac 1{10}\sin(x)-\frac 3{10}\cos(x)}$$
sorry for bad quality but the answer is correct.