Need to find the Particular Integral
$\displaystyle D^{2}y+2Dy+2y=65e^{x}sin 2x$
$\displaystyle y=\frac1{D^2+2D+2}65\mathrm e^x\sin 2x=65\mathrm e^x\frac1{(D+1)^2+2(D+1)+2}\sin 2x$ $\displaystyle =65\mathrm e^x\frac1{D^2+4D+5}\sin 2x$
$\displaystyle {}=65\mathrm e^x\frac1{-2^2+4D+5}\sin 2x=65\mathrm e^x\frac1{4D+1}\sin 2x=65\mathrm e^x\frac{4D-1}{16D^2-1}\sin 2x$
$\displaystyle {}=65\mathrm e^x\frac{4D-1}{16(-2^2)-1}\sin 2x=\mathrm e^x(1-4D)\sin 2x=\mathrm e^x(\sin 2x-8\cos 2x)$.