Solve the equation $(D^2+2D+1)y=x\cos x$ where $D=\frac {d}{dx}$
My Attempt:
The given equation is
$$(D^2+2D+1)y=x\cos x$$
It's auxiliary equation is
$$m^2+2m+1=0$$
$$(m+1)^2=0$$
$$m=-1,-1$$
$$\textrm {Complementary Function (C.F)}=(c_1+c_2 x)e^{-x}$$
Now, the particular integral is
$$\textrm {P.I}=\frac {x\cos x}{D^2+2D+1}$$
$$=x\cdot \frac {\cos x}{D^2+2D+1} – \frac {2D+2}{(D^2+2D+1)^2}\cdot \cos x$$
$$=x\cdot \frac {\cos x}{-1+2D+1} – \frac {2(D+1)}{(D+1)^4} \cdot \cos x$$
$$=x\cdot \frac {\cos x}{2D} – 2 \frac {\cos x}{(D+1)^3}$$
How to solve further?
Best Answer
Taking from where you left $$P.I=x\cdot \frac {\cos x}{2D} - 2 \frac {\cos x}{(D+1)^3}$$ $$=\frac{x\sin x}{2}-2\frac{(D-1)^3}{(D^2-1)^3}\cos x$$ $$=\frac{x\sin x}{2}-\frac{2}{-8}(D^3-3D^2+3D-1)\cos x$$ $$=\frac{x\sin x}{2}+\frac{1}{4}[\sin x+3\cos x-3\sin x -\cos x]$$ $$=\frac{x\sin x}{2}+\frac{1}{2}[\cos x-\sin x]$$