Differential equation via Laplace transform $$xy''+(2x+3)y'+(x+3)y=3e^{-x}$$
$$L[xy'']+L[(2x+3)y']+L[(x+3)y]=L[3e^{-x}]$$
$$-\frac{d}{dp}(p^2Y)-\frac{d}{dp}(2pY)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2)-\frac{dY}{dp}(2p)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2+2p+1)=\frac{3}{p+1}$$
$$-dY=-\frac{3dp}{p^3+3p^2+3p+1}$$
$$\int-dY=\int\frac{3dp}{p^3+3p^2+3p+1}$$
$$-Y=\frac{-3}{2(x+1)^2}+C$$
$$Y=\frac{3}{2(x+1)^2}+C$$
This is the answer which I got, however, at the end of the book it says that the answer is $$Y=xe^{-x}$$
Can someone please tell me if my answer is correct, and if so how I can make it look like $Y=xe^{-x} \text{?}$

Consider the equation
$$t \, y'' + (2 \, t + 3) \, y' + (t+3) \, y = a \, e^{-t}$$ 
then by the standard Laplace transform this becomes
\begin{align}
- \partial_{s}(s^2 \, \overline{y}) + y(0) - 2 \, \partial_{s}(s \, \overline{y}) + 3 \, s \, \overline{y} - 3 \, y(0) - \overline{y}' + 3 \overline{y} &= \frac{a}{s+1} 
\end{align}
\begin{align}
- (s+1)^2 \, \overline{y}' + (s+1) \, \overline{y} &= \frac{a}{s+1} + 2 \, y(0) \\
\overline{y}' - \frac{1}{s+1} \, \overline{y} &= - \frac{a}{(s+1)^3} - \frac{2 \, y(0)}{(s+1)^2} \\
\frac{d}{ds} \left( \frac{\overline{y}}{s+1} \right) &= \frac{a}{3} \, \frac{d}{ds} \left( \frac{1}{(s+1)^3} \right) + 2 \, y(0) \, \frac{d}{ds} \left( \frac{1}{(s+1)^2} \right)  \\
\frac{\overline{y}}{s+1} &= \frac{a}{3} \, \frac{1}{(s+1)^3} + 2 \, y(0) \, \frac{1}{(s+1)^2} + c_{0} \\
\overline{y} &= \frac{a}{3} \, \frac{1}{(s+1)^2} + \frac{2 \, y(0)}{s+1} + c_{0} \, (s+1). 
\end{align}
Inversion of the transform leads to
$$y(t) = \frac{a}{3} \, t \, e^{-t} + 2 \, y(0) \, e^{-t} + c_{0} \, (\delta(t) + \delta'(t))$$.
For this case supposing $y(0) = 0$, $a=3$, and since $c_{0}$ is arbitrary, or that other conditions apply, then
$$y(t) = t \, e^{-t}.$$
