Let me change your notation and put $y_h(x)$ as the solution you have obtained. Then:
$$y(x) = c_1 y_1(x) + c_2 y_2(x) + y_p(x),$$
where $y_p(x)$ is the particular solution of your ode. Then make use of variation of parameters and try solutions $y(x) = C(x) y_1(x)$ (for example; you could have chosen $y = C y_2$). Substitute back in the ode and you will have:
$$x^2 (C''y_1 + 2 C' y_1' + C y_1'') - 2 C y_1 = C'' x^2 + C'(2y_1' x^2/y_1) + C(x^2 y''_1 - 2 y_1) =x^3 e^x,$$
the coefficient in $C$ must vanish since $y_1$ is a solution of the homogenous part of the equation. Simplify a bit and obtain a "false" second order differential equation for $C(x)$:
$$ C'' + C' 2y'_1/y_1 = x e^x/y_1,$$
which can be solved in terms of an integrating factor, $u = e^{\int 2y'_1/y_1 \,dx} = y_1^2$, as follows:
$$\frac{d}{dx}\left( C' u\right) = u x e^x/y_1 = y_1 x e^x \Leftrightarrow C' = \frac{1}{y_1^2} \int y_1 x e^x \, dx + \frac{A}{y_1^2},$$
and hence the solution for $C$:
$$C = B + A \int\frac{dx}{y_1^2} + \int\left(\frac{1}{y_1^2} \int y_1 x e^x \, dx\right) \, dx,$$
where $A$ and $B$ are constants of integration. This readily leads you to the solution, $y(x) = C(x) y_1$:
$$y(x) =\frac{A x^2}{3}+\frac{B}{x}+\frac{2 e^x}{x}+e^x( x -2 ),$$
where you can make $c_1 = \frac{A}{3}$ (and $c_2 = B$).
I hope this may be useful to you.
Cheers!
A particular solution is obviously a degree $1$ polynomial: $y_0=at+b$. Since $y_0'=a$ and $y_0''=0$, the relation yields $4a+at+b=t$, that is, $a=1$ and $b=-4$.
For the general solution the general method does work! When the characteristic polynomial has a root $\lambda$ of multiplicity $m$, you get linearly independent solutions of the form $t^{k}e^{\lambda t}$, for $k=0,1,\dots,m-1$.
In this case, you get $y_1=e^{-2t}$ and $y_2=te^{-2t}$.
Best Answer
The transformed ODE is $D^2y=ze^{2z}$
For the solution of corresponding homogeneous equation the solution is $y_h=az+b=aIn(x)+b$ (Integrate $D^2y=0$ twice)
For the term $ze^{2z}$ in RHS, the particular solution is given as
$y_p= \frac {1}{D^2}ze^{2z}=\int\int ze^{2z}dz$ etc.
The general solution is $y(x)=y_h+y_p$. Can you proceed further?