Exercise:
Solve the following ODE, $y''-3y'-2y=\sin \left( {{e}^{-x}} \right)$
Solution: Let $y={{e}^{\lambda x}}$ be the trial solution of $y''-3y'-2y=0$
i.e $y'=\lambda {{e}^{\lambda x}}\,\,\And \,\,y''={{\lambda }^{2}}{{e}^{\lambda x}}$
So $y''-3y'-2y=0\Leftrightarrow {{e}^{\lambda x}}\left( {{\lambda }^{2}}-3\lambda -2 \right)=0$
thus ${{\lambda }^{2}}-3\lambda -2=0\Leftrightarrow {{\lambda }_{1}}=\frac{3-\sqrt{17}}{2}\,\,\,or\,\,\,{{\lambda }_{2}}=\frac{3+\sqrt{17}}{2}\,\,$
so ${{y}_{H}}\left( x \right)={{c}_{1}}{{e}^{\frac{3-\sqrt{17}}{2}x}}+{{c}_{2}}{{e}^{\frac{3+\sqrt{17}}{2}x}}$
Let ${{y}_{p}}\left( x \right)=A\sin \left( {{e}^{-x}} \right)+B{{e}^{x}}\sin \left( {{e}^{-x}} \right)+C{{e}^{2x}}\sin \left( {{e}^{-x}} \right)$
$+D\cos \left( {{e}^{-x}} \right)+E{{e}^{x}}\cos \left( {{e}^{-x}} \right)+F{{e}^{2x}}\cos \left( {{e}^{-x}} \right)$
so $y{{'}_{p}}\left( x \right)=-A{{e}^{-x}}\cos \left( {{e}^{-x}} \right)-B\cos \left( {{e}^{-x}} \right)+B{{e}^{x}}\sin \left( {{e}^{-x}} \right)-C{{e}^{x}}\cos \left( {{e}^{-x}} \right)+2C{{e}^{2x}}\sin \left( {{e}^{-x}} \right)$
$+D{{e}^{-x}}\sin \left( {{e}^{-x}} \right)+E{{e}^{x}}\cos \left( {{e}^{-x}} \right)+E\sin \left( {{e}^{-x}} \right)+2F{{e}^{2x}}\cos \left( {{e}^{-x}} \right)+F{{e}^{x}}\sin \left( {{e}^{-x}} \right)$
$y'{{'}_{p}}\left( x \right)=A{{e}^{-x}}\cos \left( {{e}^{-x}} \right)-A{{e}^{-2x}}\sin \left( {{e}^{-x}} \right)-B{{e}^{-x}}\sin \left( {{e}^{-x}} \right)-B\cos \left( {{e}^{-x}} \right)+B{{e}^{x}}\sin \left( {{e}^{-x}} \right)$
$-C{{e}^{x}}\cos \left( {{e}^{-x}} \right)-C\sin \left( {{e}^{-x}} \right)-2C{{e}^{x}}\cos \left( {{e}^{-x}} \right)+4C{{e}^{2x}}\sin \left( {{e}^{-x}} \right)-D{{e}^{-2x}}\cos \left( {{e}^{-x}} \right)-D{{e}^{-x}}\sin \left( {{e}^{-x}} \right)$
$+E{{e}^{x}}\cos \left( {{e}^{-x}} \right)+E\sin \left( {{e}^{-x}} \right)-E{{e}^{-x}}\cos \left( {{e}^{-x}} \right)+4F{{e}^{2x}}\cos \left( {{e}^{-x}} \right)+2F{{e}^{x}}\sin \left( {{e}^{-x}} \right)$
$-F\cos \left( {{e}^{-x}} \right)+F{{e}^{x}}\sin \left( {{e}^{-x}} \right)$
So ${{y}_{G}}\left( x \right)={{y}_{H}}\left( x \right)+{{y}_{p}}\left( x \right)={{c}_{1}}{{e}^{\frac{3-\sqrt{17}}{2}x}}+{{c}_{2}}{{e}^{\frac{3+\sqrt{17}}{2}x}}+{{e}^{2x}}\sin \left( {{e}^{-x}} \right)$
Or we can use Variation of parameters & Cramer’s
We have ${{y}_{H}}\left( x \right)={{c}_{1}}{{y}_{1}}+{{c}_{2}}{{y}_{2}}$ so ${{y}_{p}}\left( x \right)={{c}_{1}}\left( x \right){{y}_{1}}+{{c}_{2}}\left( x \right){{y}_{2}}$
Observe that \[\left\{ \begin{align}
& c_{1}^{'}\left( x \right){{y}_{1}}+c_{2}^{'}\left( x \right){{y}_{2}}=0 \\
& c_{1}^{'}\left( x \right){{y}_{1}}+c_{2}^{'}\left( x \right){{y}_{2}}=g\left( x \right) \\
\end{align} \right.\] and $W\left( {{y}_{1}},{{y}_{2}} \right)=\left| \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
y_{1}^{'} & y_{2}^{'} \\
\end{matrix} \right|$ such that $W\ne 0$
Let $A\left( g,{{y}_{2}} \right)=\left| \begin{matrix}
0 & {{y}_{2}} \\
g\left( x \right) & y_{2}^{'} \\
\end{matrix} \right|\,\,\,\,\,\And \,\,\,\,B\left( {{y}_{1}},g \right)=\left| \begin{matrix}
{{y}_{1}} & 0 \\
y_{1}^{'} & g\left( x \right) \\
\end{matrix} \right|$ then
$c_{1}^{'}\left( x \right)=\frac{A\left( g,{{y}_{2}} \right)}{W}\,\,\,\,\And \,\,\,c_{2}^{'}\left( x \right)=\frac{B\left( {{y}_{1}},g \right)}{W}$
Let $\lambda =\frac{3-\sqrt{17}}{2}\,\,\,\And \,\,\beta =\frac{3+\sqrt{17}}{2}$ then ${{y}_{1}}={{e}^{\lambda x}}\,\,\,\And \,\,\,\,{{y}_{2}}={{e}^{\beta x}}$
So $W\left( {{y}_{1}},{{y}_{2}} \right)=\left| \begin{matrix}
{{e}^{\lambda x}} & {{e}^{\beta x}} \\
\lambda {{e}^{\lambda x}} & \beta {{e}^{\beta x}} \\
\end{matrix} \right|=\beta {{e}^{\left( \lambda +\beta \right)x}}-\lambda {{e}^{\left( \lambda +\beta \right)x}}={{e}^{\left( \lambda +\beta \right)x}}\left( \beta -\lambda \right)\ne 0$
Hence ${{y}_{1}}\,\,\And \,\,\,\,{{y}_{2}}$ are linearly independent
$A\left( g,{{y}_{2}} \right)=\left| \begin{matrix}
0 & {{e}^{\beta x}} \\
\sin \left( {{e}^{-x}} \right) & \beta {{e}^{\beta x}} \\
\end{matrix} \right|=-{{e}^{\beta x}}\sin \left( {{e}^{-x}} \right)$ and$B\left( {{y}_{1}},g \right)=\left| \begin{matrix}
{{e}^{\lambda x}} & 0 \\
\lambda {{e}^{\lambda x}} & \sin \left( {{e}^{-x}} \right) \\
\end{matrix} \right|={{e}^{\lambda x}}\sin \left( {{e}^{-x}} \right)$
$c_{1}^{'}\left( x \right)=\frac{-{{e}^{\beta x}}\sin \left( {{e}^{-x}} \right)}{\left( \beta -\lambda \right){{e}^{\left( \lambda +\beta \right)x}}}=\frac{{{e}^{-\lambda x}}\sin \left( {{e}^{-x}} \right)}{\lambda -\beta }$ and $c_{2}^{'}\left( x \right)=\frac{{{e}^{-\beta x}}\sin \left( {{e}^{-x}} \right)}{\beta -\lambda }$
So ${{y}_{p}}\left( x \right)=\frac{{{e}^{\lambda x}}}{\lambda -\beta }\int{{{e}^{-\lambda x}}\sin \left( {{e}^{-x}} \right)dx}+\frac{{{e}^{\beta x}}}{\beta -\lambda }\int{{{e}^{-\beta x}}\sin \left( {{e}^{-x}} \right)dx}$
But $\int{{{e}^{-\lambda x}}\sin \left( {{e}^{-x}} \right)dx}=\int{{{\left( {{e}^{-x}} \right)}^{\lambda }}\sin \left( {{e}^{-x}} \right)dx}$
Let $u={{e}^{-\left( \lambda -1 \right)x}}\,\,\And \,\,\,dv=\int{{{e}^{-x}}\sin \left( {{e}^{-x}} \right)dx}\Leftrightarrow du=-\left( \lambda -1 \right){{e}^{-\left( \lambda -1 \right)x}}dx\,\,\And \,\,v=\cos \left( {{e}^{-x}} \right)$
So $\int{{{e}^{-\left( \lambda -1 \right)x}}}{{e}^{-x}}\sin \left( {{e}^{-x}} \right)dx=\cos \left( {{e}^{-x}} \right){{e}^{-\left( \lambda -1 \right)x}}-\left( \lambda -1 \right)\int{\cos \left( {{e}^{-x}} \right)}\,{{e}^{-\left( \lambda -1 \right)x}}dx$
Also $\int{\cos \left( {{e}^{-x}} \right){{e}^{-\left( \lambda -1 \right)x}}dx}=\int{{{e}^{-\left( \lambda -2 \right)x}}{{e}^{-x}}\cos \left( {{e}^{-x}} \right)dx}$
Let $u={{e}^{-\left( \lambda -2 \right)x}}\,\,\And \,\,\,dv=\int{{{e}^{-x}}\cos \left( {{e}^{-x}} \right)dx}\Leftrightarrow du=-\left( \lambda -2 \right){{e}^{-\left( \lambda -2 \right)x}}dx\,\,\And \,\,v=-\sin \left( {{e}^{-x}} \right)$
So $\int{\cos \left( {{e}^{-x}} \right){{e}^{-\left( \lambda -1 \right)x}}dx}=-\sin \left( {{e}^{-x}} \right){{e}^{-\left( \lambda -2 \right)x}}-\left( \lambda -2 \right)\int{\sin \left( {{e}^{-x}} \right){{e}^{-\left( \lambda -2 \right)x}}dx}$
*____________________________________
first Method Credit to Katrine Linda
No comments:
Post a Comment