Question

Solve $(D^2+4D+3)y=e^{-x}\sin x+x{e}^{3x}$.

Answer 1

$(D+1)(D+3)y=e^{-x}\sin x+x{e}^{3x}$

C.F. = $C_1{e}^{-x}+C_2{e}^{-3x}$

For P.I. , we get

$\frac{e^{-x}\sin x+x{e}^{3x}}{D^2+4D+3}$

$e^{-x}\frac{\sin x}{D(D+2)}+e^{3x}\frac{x}{(D+4)(D+6)}$

$e^{-x}\frac{(D-2)\sin x}{D(D^2-4)}+e^{3x}\frac{x}{24\frac{(D^2+10D)}{24}+1}$

$=e^{-x}\frac{(\sin x+2\cos x)}{-3}+e^{3x}\frac{5}{288}$