D operators method

Question

Find P.I. of the following equation: $$(D^2-1)y=(1+x^2)e^x$$ I have tried like this: $$\begin{align} P.I.& =\frac{(1+x^2)e^x}{D^2-1}\\ & =e^x\frac{(1+x^2)}{(D+1)^2-1}\\ & =e^x\frac{(1+x^2)}{D^2+2D}\\ & =e^x\frac{1}{D+2}\frac{1}{D}(1+x^2)\\ & =e^x\frac{1}{D+2}\left(x+\frac{x^3}{3}\right)\\ & =e^x\frac{1}{D+2}x+\frac{e^x}{3}\frac{1}{D+2}x^3\\ & =.....................\\ & =\frac{1}{2}xe^x+\frac{1}{6}x^3e^x-\frac{3}{8}e^x\\ \end{align}$$ But in my book the answer is: $$\frac{1}{12}xe^x(2x^2-3x+9)$$ Please check where is my fault

Answer 1

Let us start with OP's work from the step: $$y=e^x\frac{1}{D+2} x+ \frac{e^x}{3}\frac{1}{D+2}x^3$$ $$\implies y=\frac{e^x}{2}(1+D/2)^{-1} x+\frac{e^x}{6}(1+D/2)^{-1}$$ $$\implies y=\frac{e^x}{2}(1-D/2+D^2/4+..)x+\frac{e^x}{6}(1-D/2+D^2/4-D^3/8)x^3$$ $$\implies y=e^x[x^3/6-x^2/4+3x/4-3/8]~~~(1)$$

Hence OP's and book's answers are not correct. The simple indicator in OP's answer is the absence of $x^2$. Absence of constant term in book's answer indicates error. Using $y$ from (1) one can check that $$(D^2-1)y=(1+x^2)e^x$$