Find the general solution of: $y^"+3y^{'}+2y=\sin(x)$ (I'm using the Annihilator Method)

Question

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.

Answer 1

sorry for bad quality but the answer is correct.

Answer 2

$$y^"+3y^{'}+2y=\sin(x)$$ Using annihilator 's method $$(D+1)(D+2)(D^2+1)y=0$$ $$ \begin{align} y_h=& \, c_1e^{-x}+c_2e^{-2x} \\ y_p=& \, C\sin(x)+D\cos(x) \end{align} $$ $$(-C\sin(x)-D\cos(x)+3(C\cos(x)-D\sin(x))+2(C\sin(x)+D\cos(x))=\sin(x)$$ $$ \begin{cases} -D+3C+2D=0 \\ -C-3D+2C=1 \end{cases} \implies \begin{cases} D=-3C \\ C-3D=1 \end{cases} \implies \begin{cases} D=-3/10\\ C=1/10 \end{cases} $$ $$\boxed{y_p=\frac 1{10}\sin(x)-\frac 3{10}\cos(x)}$$