Solving $(D^2-1)y=e^x(1+x)^2$

Question

I did like this: $$\text{Let,} y=e^{mx} \text{ be a trial solution of } (D^2-1)y=0$$ $$\therefore \text{The auxiliary equation is } m^2-1=0$$ $$\therefore m=\pm1\\ \text{C.F.} = c_1e^x+c_2e^{-x}$$ $$\begin{align} \text{P.I.}& =\frac{1}{D^2-1}e^x(1+x)^2\\ & =e^x\frac{1}{(D+1)^2-1}(1+2x+x^2)\\ & =e^x\frac{1}{D^2+2D}(1+2x+x^2)\\ & =\frac{e^x}{2}\left[\frac{1}{D}-\frac{1}{D+2}\right](1+2x+x^2)\\ & =\begin{aligned} \frac{e^x}{2}\frac{1}{D}(1+2x+x^2)-\frac{e^x}{2}\frac{1}{D+2}(1+2x& +x^2)\\ \end{aligned}\\ & =\frac{e^x}{2}\left(x+x^2+\frac{x^3}{3}\right)-\frac{e^x}{4}\left(x^2+x+\frac{1}{2}\right)\\ & =e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)\\ \end{align}$$ $$\therefore \text{The solution is}$$ $$y=c_1e^x+c_2e^{-x}+e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)$$ But, in my book the answer is: $$y=c_1e^x+c_2e^{-x}+\frac{xe^x}{12}(2x^2+3x+3)$$ $$\text{Please, check if there is any } \color{red}{mistake}.$$

Answer 1

The answers are essentially the same. $-\frac18e^x$ can be absorbed in $c_1e^x$.

Answer 2

It is the same. Your coefficient for $e^x$ is $c_1-\frac 18$, which you can rename it to say $c_3$