Given : $\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+y=2e^{3x}$
The auxiliary equation is
$D^{2} + 2D + 1 = 0 or m^{2} + 2m + 1 = 0$
$\Rightarrow \left(m+1\right)\left(m+1\right) = 0 \Rightarrow m =-1, -1$
i.e., repeated roots
$\therefore $ Complementary function = $\left(c_{1} + c_{2}x\right)e^{-x}$
Now Particular Integral $\left(P.I.\right)$
$=\frac{1}{D^{2}+2D+1}.2e^{3x}\quad\quad\quad\quad\quad\quad\left[D = 3\right]$
P.I$=\frac{1}{3^{2}+2.3+1}.2.e^{3x}=\frac{2e^{3x}}{16}=\frac{e^{3x}}{8}$
Solution y = C. F. + P. I.
$\Rightarrow y=\left(c_{1} + c_{2}x\right)e^{-x}+\frac{e^{3x}}{8}$