2.6.2.1 Example 1 \(x^{2}y^{\prime \prime }+xy^{\prime }+y=1\)

\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=1 \]

Comparing the ode to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {1}{x^{2}}\). There is one singular point at \(x_{0}=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}1=1\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r+1 & =0\\ r^{2}+1 & =0\\ r & =\pm i \end{align*}

Hence \(r_{1}=i,r_{2}=-i\). Expansion around \(x=0\). This is regular singular point.  Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

Solving first for the homogenous ode.

\begin{align*} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

For \(n=0\)

\begin{equation} \left ( r\left ( r-1\right ) +r+1\right ) a_{0}x^{r}=0 \tag {1}\end{equation}

Since \(a_{0}\neq 0\), then \(\left ( r\left ( r-1\right ) +r+1\right ) =0\) or \(\allowbreak r^{2}+1=0\). Therefore \(r=\pm i\) as was found above. The homogenous ode therefore satisfies

\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=\left ( r^{2}+1\right ) a_{0}x^{r}\]

Since when \(r=\pm i\), the RHS is zero. For \(n\geq 1\) the recurrence relation is

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n} & =0\nonumber \\ \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) +1\right ) a_{n} & =0\nonumber \\ \left ( n^{2}+2nr+r^{2}+1\right ) a_{n} & =0 \tag {2}\end{align}

Let \(a_{0}=1\). For \(r=i\).  For \(n=1\)

\[ \left ( 1+2i-1+1\right ) a_{1}=0 \]

Hence \(a_{1}=0\). Similarly all \(a_{n}=0\) for \(n\geq 1\). Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+i}\\ & =x^{i}\left ( a_{0}+a_{1}x+\cdots \right ) \\ & =a_{0}x^{i}\\ & =x^{i}\end{align*}

For \(r=-i\).  For \(n=1\) and using \(b\) instead of \(a\), we obtain (also using \(b_{0}=1\))

\[ \left ( 1-2i+1+1\right ) b_{n}=0 \]

Hence \(b_{1}=0\). Similarly all \(b_{n}=0\) for \(n\geq 1\). Hence

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-i}\\ & =x^{-i}\left ( b_{0}+b_{1}x+\cdots \right ) \\ & =b_{0}x^{-i}\\ & =x^{-i}\end{align*}

Therefore

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x^{i}+c_{2}x^{-i}\end{align*}

To find \(y_{p}\) since the ode satisfies

\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=\left ( r^{2}+1\right ) a_{0}x^{r}\]

Relabel \(r=m,a_{0}=c_{0}\) to avoid confusion with terms used above,  then we balance RHS, hence

\[ \left ( m^{2}+1\right ) c_{0}x^{m}=1 \]

This implies \(m=0\) and \(c_{0}=1\). Therefore

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

Using the recurrence relation (2) found above, but now using the values found \(m=0\) and \(c_{0}=1\), then (2) becomes

\begin{align*} \left ( n^{2}+2nm+m^{2}+1\right ) c_{n} & =0\\ \left ( n^{2}+1\right ) c_{n} & =0 \end{align*}

Hence all \(c_{n}=0\) except for \(c_{0}\). Therefore

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}\\ & =1 \end{align*}

Hence the solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{i}+c_{2}x^{-i}+1 \end{align*}