Find a particular solution of $\,\,y''+3y'+2y=\exp(\mathrm{e}^x)$ I already solved for the homogeneous one, but I'm still looking for the particular solution of the differential equation:
$$y''+3y'+2y=\exp(\mathrm{e}^x)$$
The homogeneous solutions of this system are $\mathrm{e}^{-x}$ and $\mathrm{e}^{-2x}$. I've tried the substitution 
$$y_p(x)=v(x)\mathrm{e}^{-2x},$$ which yielded the differential equation:
$$v''+v'=\exp(\mathrm{e}^x+2x),$$
After that I reduced the order by $p=v'$ which gives:
$$p'+p=\exp(\mathrm{e}^x+2x).$$
Now this equation still seems hard to solve. I was wondering if there were easier/better substitutions to make ?
 A: We first observe that
$$
y=\frac{1}{(n+1)(n+2)}\mathrm{e}^{nx},
$$
is a particular solution of
$$
y''+3y'+2y=\mathrm{e}^{nx}.
$$
Hence
$$
\sum_{n=0}^\infty \frac{1}{(n+1)(n+2)n!}\mathrm{e}^{nx}=\mathrm{e}^{-2x}\left(\exp(\mathrm{e}^x)-1-\mathrm{e}^x\right)
$$
is a particular solution of 
$$
y''+3y'+2y=\sum_{n=0}^\infty\frac{1}{n!}\mathrm{e}^{nx}=\exp(\mathrm{e}^x).
$$
Note. However, if one wants to be rigorous, a verification would be the right thing to do.
A: This solution might be more time-comsuming, but it can work.
We can use a method called variation of parameters.
If we take our characteristic polynomial:
$$y_c(t) = r^2+3r+2 = (r+2)(r+1)$$
Now we have that our complementary solution is:
$$y_c(t) = c_1e^{-2t} + c_2e^{-t}$$
Now our solution to this equation is $e^{-2t}$ and $e^{-t}$
We can use this to find our Wronskian
$$W = \begin{vmatrix}e^{-2t} &e^{-t} \\
-2e^{-2t} & -e^{-t} \end{vmatrix} = -e^{-3t}+2e^{-3t} = e^{-3t}$$
Now we can find our particular solution.
$$y_p(t) = -e^{-2t}\int\frac{e^{-t}e^{e^t}}{e^{-3t}}dt + e^{-t}\int\frac{e^{-2t}e^{e^t}}{e^{-3t}}dt$$
Solving this will get you the particular solution.
EDIT
Solving for the particular we get 
$$y_p(t) = -e^{-2t}\left(e^{e^t}(e^{t}-1)\right) + e^{-t}e^{e^t}$$
$$y_p(t) = -e^{-2t}(e^{e^t}e^{t}-e^{e^t}) + e^{-t}e^{e^t} =  e^{-2t}e^{e^t}$$
If you need the general solution, add both the complementary and the particular solution together.
A: I believe Claude's comment is correct an that differential equation is not pretty.  So I'll attempt a different approach, attempting to solve for the general solution.
Since the characteristic polynomial is $(s+2)(s+1)$, it appears integrating factors of $e^x$ or $e^{2x}$ will work.  $e^x$ will result in easier integration.  So we let
$$z=y'+2y$$
$$y''+3y'+2y=(y'+2y)'+y'+2y=z'+z=e^{e^x}$$
$$e^xz'+e^xz=(e^xz)'=e^xe^{e^x}$$
$$e^xz=e^{e^x}+k_1$$
$$e^x(y'+2y)=e^{e^x}+k_1$$
$$e^{2x}(y'+2y)=(e^{2x}y)'=e^xe^{e^x}+k_1e^x$$
$$e^{2x}y=e^{e^x}+k_1e^x+k_2$$
$$y=e^{e^x-2x}+k_1e^{-x}+k_2e^{-2x}$$
