Second order linear differential equations : Particular Integral 
How do they find the particular Integral. These two lines are not giving me any clue. Please someone explain me.

 A: For my own comfort, I will use $t$ instead of $x$. 
Let me try to explain why
\begin{align}
\frac{1}{(D+4)(D+2)}e^{-2t}= \frac{te^{-2t}}{\theta(-2)} = \frac{1}{2}te^{-2t}.
\end{align}
We shall prove the following identity

\begin{align} p(D)[te^{zt}] = p(z)te^{zt} + p'(z) e^{zt}  \end{align}

when $P(D)$ is quadratic (you could prove it for any polynomial $p(z)$). Observe
\begin{align}
\frac{d}{dt}[te^{zt}]= e^{zt}+zte^{zt} \ \ \text{ and } \ \ \frac{d^2}{dt^2}[te^{zt}] = 2ze^{zt}+z^2te^{zt}
\end{align} 
which means
\begin{align}
p\left( D\right)[te^{zt}]=&\  \left(a_2\frac{d^2}{dt^2}+a_1\frac{d}{dt} + a_0 \right)te^{zt}\\\
 =&\ a_2(2ze^{zt}+z^2te^{zt})+a_1(e^{zt}+zte^{zt}) + a_0te^{zt} \\
=&\ (a_2z^2+a_1z+a_0)te^{zt} +(2a_2z+a_1)e^{zt}\\
=&\ p(z)te^{zt} + p'(z)e^{zt}. 
\end{align}
Using the above identity, you see that
\begin{align}
p(D)[te^{-2t}] = p(-2)te^{-2t}+p'(-2)e^{-2t} = p'(-2) e^{-2t}
\end{align}
since $z=-2$ is a root of the polynomial $p(z)$. Hence you have the identity
\begin{align}
\frac{1}{p'(-2)}p(D)[te^{-2t}]=p(D)\left[\frac{te^{-2t}}{p'(-2)} \right]= e^{-2t}.
\end{align}
Your book basicially called $p'(z) = \theta(z)$. 
In conclusion, since
\begin{align}
p(D)\left[\frac{te^{-2t}}{p'(-2)} \right]= e^{-2t}
\end{align}
then symbolically you have

\begin{align} 
\frac{1}{p(D)}e^{-2t}=\frac{te^{-2t}}{p'(-2)}. \end{align}

Additional note: since $p(z) = (z+2)(z+4)$ then $p'(z) = (z+4) +(z+2)$. 
A: The method of variation of parameters, however, provides one extra term.
$y_p=-\frac{1}{4}\mathrm{e}^{-2x}+\frac{x}{2}\mathrm{e}^{-2x}$
Solution Procedure:
Two fundamental solutions are $y_1(x)=\mathrm{e}^{-2x}$ and $y_2(x)=\mathrm{e}^{-4x}$.
Thus, $y_p=y_2(x)\int\frac{y_1(x)f(x)}{W(y_1,y_2)}\mathrm{d}x-y_1(x)\int\frac{y_2(x)f(x)}{W(y_1,y_2)}\mathrm{d}x$
where, $W(y_1,y_2)=y_1(x)y_2'(x)-y_1'(x)y_2(x)$ and $f(x)=\mathrm{e}^{-2x}$.
