ODE Equation solved by using Cauchy-Euler Method asked by Katrina Linda in the pretty math group

Exercise:

Solve the D.E ${{\left( x-3 \right)}^{2}}y''-\left( x-3 \right)y'+y=4x+7$

Solution: let $u=x-3\Leftrightarrow x=u+3$

So the D.E transform to the form ${{u}^{2}}y''-uy'+y=4u+19$  (*)

Observe that ${{u}^{2}}y''-uy'+y=0$ is an Cauchy Euler homogenous D.E

Suppose that $y={{u}^{\lambda }}$ is the solution of the above homogenous differential equation

So $y'=\lambda {{u}^{\lambda -1}}\,\,\And \,\,y''=\lambda \left( \lambda -1 \right){{u}^{\lambda -2}}=\left( {{\lambda }^{2}}-\lambda  \right){{u}^{\lambda -2}}$ substitute to get

${{u}^{2}}\left( {{\lambda }^{2}}-\lambda  \right){{u}^{\lambda -2}}-u\lambda {{u}^{\lambda -1}}+{{u}^{\lambda }}=0\Leftrightarrow \left( {{\lambda }^{2}}-\lambda  \right){{u}^{\lambda }}-\lambda {{u}^{\lambda }}+{{u}^{\lambda }}=0$

$\Leftrightarrow {{u}^{\lambda }}\left( {{\lambda }^{2}}-\lambda -\lambda +1 \right)=0\Leftrightarrow {{u}^{\lambda }}=0\,\,or\,{{\lambda }^{2}}-2\lambda +1=0$

So ${{\left( \lambda -1 \right)}^{2}}=0\Leftrightarrow {{\lambda }_{1}}={{\lambda }_{2}}=1$ (double root)

So ${{y}_{H}}\left( u \right)={{c}_{1}}{{u}^{1}}+{{c}_{2}}{{u}^{1}}\ln u=u\left( {{c}_{1}}+{{c}_{2}}\ln u \right)$

Hence ${{y}_{H}}\left( x \right)=\left( x-3 \right)\left( {{c}_{1}}+{{c}_{2}}\ln \left( x-3 \right) \right)$

Now to find the particular solution we will use variation parameter method

We have ${{u}^{2}}y''-uy'+y=4u+19\Leftrightarrow y''-\frac{1}{u}y'+\frac{1}{{{u}^{2}}}y=\frac{4}{u}+\frac{19}{{{u}^{2}}}$

Since $\frac{1}{u}\,\,,\,\,\frac{1}{{{u}^{2}}}\,\,\,\And \,\,\,\frac{4}{u}+\frac{19}{{{u}^{2}}}$ are continuous on the interval and ${{y}_{1}}\left( u \right)=u\,\,\And \,\,{{y}_{2}}\left( u \right)=u\ln u$

are linearly independent so the particular solution will be

 ${{y}_{p}}\left( u \right)=-{{y}_{1}}\int{\frac{{{y}_{2}}g\left( u \right)}{W\left( {{y}_{1}},{{y}_{2}} \right)}du}+{{y}_{2}}\int{\frac{{{y}_{1}}g\left( u \right)}{W\left( {{y}_{1}},{{y}_{2}} \right)}du}$

So $W\left( {{y}_{1}},{{y}_{2}} \right)=\left| \begin{matrix}
   u & u\ln u  \\
   1 & 1+\ln u  \\
\end{matrix} \right|=u\left( 1+\ln u \right)-u\ln u=u+u\ln u-u\ln u=u$

So ${{y}_{p}}\left( u \right)=-u\int{\frac{u\ln u\left( \frac{4u+19}{{{u}^{2}}} \right)}{u}du}+u\ln u\int{\frac{u\left( \frac{4u+19}{{{u}^{2}}} \right)}{u}du}$

Notice that , $\frac{\frac{4u+19}{u}}{u}=\frac{4u+19}{{{u}^{2}}}=\frac{4}{u}+\frac{19}{{{u}^{2}}}$

So $\int{\frac{4u+19}{{{u}^{2}}}du}=\int{\frac{4}{u}du}+\int{\frac{19}{{{u}^{2}}}du}=4\ln u-\frac{19}{u}$

$\Rightarrow u\ln u\int{\frac{4u+19}{{{u}^{2}}}du}=u\ln u\left( 4\ln u-\frac{19}{u} \right)=4u{{\ln }^{2}}u-19\ln u$

Also $\int{\frac{u\ln u\left( \frac{4u+19}{{{u}^{2}}} \right)}{u}du}=\int{\ln u\left( \frac{4}{u}+\frac{19}{{{u}^{2}}} \right)du}=4\int{\frac{\ln u}{u}du}+19\int{\frac{\ln u}{{{u}^{2}}}du}$

                                  $=2{{\ln }^{2}}u-\frac{19}{u}-\frac{19\ln u}{u}$

So $-u\int{\frac{u\ln u\left( \frac{4u+19}{{{u}^{2}}} \right)}{u}du}=-u\left( 2{{\ln }^{2}}u-\frac{19}{u}-\frac{19\ln u}{u} \right)=-2u{{\ln }^{2}}u+19\ln u+19$

Thus ${{y}_{G}}\left( u \right)={{y}_{H}}\left( u \right)+{{y}_{p}}\left( u \right)=u\left( {{c}_{1}}+{{c}_{2}}\ln u \right)+2u{{\ln }^{2}}u+19$

So ${{y}_{G}}\left( x \right)=\left( x-3 \right)\left( {{c}_{1}}+{{c}_{2}}\ln \left( x-3 \right) \right)+2\left( x-3 \right){{\ln }^{2}}\left( x-3 \right)+19$     Q.E.D