Solution :

Solving D.E equation by method of variation of parameter.

$\therefore(D^2+3D+2)y=e^{e^x} $

The auxiliary equation is

$m^2+3m+2=0 \\ \implies m=-1,-2 $

Finding Complementary (C.F) functions

$C.F=C_1y_1+C_2y_2\\ C.F = C_1e^{-2x} +C_2e^{-x}$

Where ,

$y_1= e^{-2x}, y_2=e^{-x} $

We assume particular integral (P.I)

$P.I=uy_1+vy_2$

$W=\left|\begin{array}{cc}y_1, & y_2 \\ y_1' & y_2'\end{array}\right|=\left|\begin{array}{cc}e^{-2 x} & e^{-x} \\ -2 e^{-2 x} & -e^{-x}\end{array}\right|$

$\space\space\space\space=-e^{-x} e^{-2 x}-\left(-2 e^{-2 x} e^{-x}\right)\\ \space\space \space\space=- e^{-3 x}-\left(-2 e^{-3 x}\right)\\ \space\space\space\space=e^{-3 x}$

We assumed P.I

$\begin{aligned} P.I&= uy_{1}+vy_{2} \\ &=e^{-2 x}u+e^{-x}v \end{aligned}$

$$\begin{aligned} u &=-\int \frac{y_{2} x}{W} d x \\ &=-\int \frac{e^{-x} e^{ x}}{e^{-3 x}} d x\\ &=-\int e^{-x} e^{3 x} e^{e^{x}} d x \\ &=-\int e^{2 x} e^{e^{x}} d x\\ \text{put} e^{x}=t \\ e^xdx=dt\\ &=-\int te^{t} d t \\ &=-\left[t e^{t}-e^{t}\right] \\ &=-t e^{t}+e^{t} \\ u &=-e^{x} e^{e^{x}}+e^{e^{x}} \end{aligned}$$

And

$$\begin{aligned} v &= \int \frac{y_1x}{W} d x \\ &= \int \frac{e^{-2 x} e^{e^{x}}}{e^{-3 x}} d x \\ &= \int e^{-2x} e^{3 x} e^{x} d x \\ &= \int e^{x} e^{e^{x}} d x \\ &= \int e^{t} d t \\ &=e^t \\ v &=e^{e^x} \end{aligned}$$

$\begin{aligned} \therefore P.I &=(-e^xe^{e^x}+e^{e^x})e^{-2x}+e^{e^x}e^{-x}\\ &=e^{-2x} e^{e^x} \end{aligned}$

General Solution (C.S) = C.F + P.I

$ C.S=C_1e^{-2x} +C_2e^{-x}+e^{-2x} e^{e^x} $