2nd Order ODE Solved by using Variation of Parameters to find its particular solution

Exercise:

Solve the following ODE, $y''+3y'+2y=\sin \left( {{e}^{x}} \right)$

Solution: Let $y\left( x \right)={{e}^{\lambda x}}\,,\lambda \in \mathbb{R}$ be the trial solution of $y''+3y'+2y'=0$

So $y'\left( x \right)=\lambda {{e}^{\lambda x}}\,\,\,\And \,\,\,\,y''\left( x \right)={{\lambda }^{2}}{{e}^{\lambda x}}$ hence ${{e}^{\lambda x}}\left( {{\lambda }^{2}}+3\lambda +2 \right)=0$

Thus ${{\lambda }^{2}}+3\lambda +2=0\Leftrightarrow \lambda =-1\,\,or\,\,\lambda =-2$

Thus ${{y}_{H}}\left( x \right)={{c}_{1}}{{e}^{-x}}+{{c}_{2}}{{e}^{-2x}}$

Observe that $W\left( {{y}_{1}},{{y}_{2}} \right)=\left| \begin{matrix}
   {{e}^{-x}} & {{e}^{-2x}}  \\
   -{{e}^{-x}} & -2{{e}^{-2x}}  \\
\end{matrix} \right|=-2{{e}^{-3x}}+{{e}^{-3x}}=-{{e}^{-3x}}\ne 0$ as ${{e}^{-3x}}>0$

Hence ${{y}_{1}}\,\And \,\,{{y}_{2}}$ are linearly independent

So ${{y}_{p}}\left( x \right)=-{{y}_{1}}\int{\frac{{{y}_{2}}g\left( x \right)}{W}dx}+{{y}_{2}}\int{\frac{{{y}_{1}}g\left( x \right)}{W}dx}$

            $=-{{e}^{-x}}\int{\frac{{{e}^{-2x}}\sin \left( {{e}^{x}} \right)}{-{{e}^{-3x}}}dx+{{e}^{-2x}}\int{\frac{{{e}^{-x}}\sin \left( {{e}^{x}} \right)}{-{{e}^{-3x}}}dx}}$

           $={{e}^{-x}}\int{{{e}^{-2x}}{{e}^{3x}}\sin \left( {{e}^{x}} \right)dx}-{{e}^{-2x}}\int{{{e}^{-x}}{{e}^{3x}}\sin \left( {{e}^{x}} \right)dx}$

           $={{e}^{-x}}\int{{{e}^{x}}\sin \left( {{e}^{x}} \right)dx}-{{e}^{-2x}}\int{{{e}^{2x}}\sin \left( {{e}^{x}} \right)dx}$

           $={{e}^{-x}}\int{\sin \left( {{e}^{x}} \right)d\left( {{e}^{x}} \right)-{{e}^{-2x}}\int{{{e}^{2x}}\sin \left( {{e}^{x}} \right)dx}}$

           $=-{{e}^{-x}}\cos \left( {{e}^{x}} \right)-{{e}^{-2x}}\int{{{e}^{2x}}\sin \left( {{e}^{x}} \right)dx}$

Let $u={{e}^{x}}\,\,\And \,\,dv=\int{{{e}^{x}}\sin \left( {{e}^{x}} \right)dx}=\int{\sin \left( {{e}^{x}} \right)d\left( {{e}^{-x}} \right)}$

Then $du={{e}^{x}}\,\,\,\And \,\,v=-\cos \left( {{e}^{-x}} \right)$

So $\int{{{e}^{2x}}\sin \left( {{e}^{x}} \right)dx}=-{{e}^{x}}\cos \left( {{e}^{x}} \right)+\int{{{e}^{x}}\cos \left( {{e}^{x}} \right)dx=-{{e}^{x}}\cos \left( {{e}^{x}} \right)+}\sin \left( {{e}^{x}} \right)+c$

Hence ${{y}_{p}}\left( x \right)=-{{e}^{-x}}\cos \left( {{e}^{x}} \right)-{{e}^{-2x}}\left( -{{e}^{x}}\cos \left( {{e}^{x}} \right)+\sin \left( {{e}^{x}} \right) \right)=-{{e}^{-2x}}\sin \left( {{e}^{x}} \right)$

Therefore, ${{y}_{G}}\left( x \right)={{y}_{H}}\left( x \right)+{{y}_{p}}\left( x \right)={{c}_{1}}{{e}^{-x}}+{{c}_{2}}{{e}^{-2x}}-{{e}^{-2x}}\sin \left( {{e}^{x}} \right)$