Find the solution to cauchys problem I'm learning differential equation and not sure how to solve the following problem:

Find the solution for the cauchy's problem:
$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$
$y(0)=0$
$y'(0)=1$

What I tried so far:
$\lambda^2-6\lambda+9=0 \Leftrightarrow (\lambda-3)(\lambda-3)=0$
$y(x)=c_1*e^{3x}+c_2*x*e^{3x}+\varphi(x)$
I'm not sure how to continue.
Could you help me?
Thanks
 A: To begin with, I would notice that
\begin{align*}
y'' - 6y' + 9y & = (y'' - 3y') - (3y' - 9y) = (y' - 3y)' - 3(y' - 3y)
\end{align*}
Thus if we make the change of variable $u = y' - 3y$, one gets that
\begin{align*}
u' - 3u = e^{3x}\left(\frac{x^{2}+2}{x^{2}+1}\right) & \Longleftrightarrow (ue^{-3x})' = \frac{x^{2}+2}{x^{2}+1} = 1 + \frac{1}{x^{2}+1}\\\\
& \Longleftrightarrow ue^{-3x} = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow (ye^{-3x})' = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow ye^{-3x} = \frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\\\\
& \Longleftrightarrow y(x) = e^{3x}\left(\frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\right)
\end{align*}
A: hint
It is easier to begin by putting
$$\boxed{y=z.e^{3x}}$$
with
$$y'=(z'+3z)e^{3x}$$
$$y''=(z''+6z'+9z)e^{3x}$$
thus
$$y''-6y'+9y=$$
$$\color{red}{z''}e^{3x}=\color{red}{(1+\frac{1}{1+x^2})}e^{3x}$$
and
$$z'=c_2+x+\arctan(x)$$
$$z=(c_1+c_2x)+\frac{x^2}{2}+x\arctan(x)-\frac 12\ln(1+x^2)$$
With initial conditions, you will find $$c_2=1 \;\; and \;\;c_1=0$$
A: $$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$$
The differential equation is equivalent to:
$$(ye^{-3x})''=1+\frac{1}{x^2+1}$$
Integrate twice.
