## Proof of Higher Order Inverse Differential Operator March 23

I'm having difficulty proving the following :$\frac{1}{D^2+\alpha^2}sin(\alpha x) = \frac{-x}{2\alpha}cos(\alpha x)$I know it won't use the transformation $P(D^2:\mapsto -\alpha^2)$, since it will put a zero in the denominator, and that it should inv