D operator method

Question

Verify that $y=x^2-6$ is a solution of $$y''+y'-2y=14+2x-2x^2$$ I have tried like this: $$\begin{align} P.I. &=\frac{14+2x-2x^2}{D^2+D-2}\\ & =\frac{-2(x^2-x-7)}{-2\left(1-\frac{D^2+D}{2}\right)}\\ & =x^2-x-7\left(1-\frac{D^2+D}{2}\right)^{-1}\\ & =x^2-x-7+\frac{2+2x-1}{2}\\ & =\frac{2x^2-13}{2} \end{align}$$ I can't find where is my fault, please help

Answer 1

\begin{align}(D^2+D-2)(y)&=(D^2+D-2)(x^2-6)\\&= D^2(x^2-6)+D(x^2-6)-2(x^2-6)\\&= 2+2x-2x^2+12\\&= 14+2x-2x^2 \end{align}