Example 4.4.1.
Solve
\begin{equation*}
2x^{2}y'' -xy' +(x^{2}+1)y =0.
\end{equation*}
Solution.
\begin{equation}
2x^{2}y'' -xy' +(x^{2}+1)y =0 \tag{4.4.1}
\end{equation}
This equation has a regular singularity at \(x=0\text{.}\) Hence we assume the solution in the form
\begin{equation}
y = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k} \tag{4.4.2}
\end{equation}
or,
\begin{equation*}
y' = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}(\lambda+k)x^{\lambda+k-1}
\end{equation*}
and
\begin{equation*}
y'' = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}(\lambda+k)(\lambda+k-1)x^{\lambda+k-2}
\end{equation*}
Substituting these values in eqn. (4.4.1), we get -
\begin{equation*}
2x^{2}\sum\limits_{\lambda =0}^{\infty}a_{\lambda}(\lambda+k)(\lambda+k-1)x^{\lambda+k-2}-x\sum\limits_{\lambda =0}^{\infty}a_{\lambda}(\lambda+k)x^{\lambda+k-1}
\end{equation*}
\begin{equation*}
+(x^{2}+1)\sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k} =0
\end{equation*}
or,
\begin{equation*}
\sum\limits_{\lambda =0}^{\infty}a_{\lambda}\left[2(\lambda+k)(\lambda+k-1)x^{\lambda+k-1}-(\lambda+k)x^{\lambda+k}+x^{\lambda+k}\right]
\end{equation*}
\begin{equation*}
+\sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k+2} =0
\end{equation*}
or,
\begin{equation*}
\sum\limits_{\lambda =0}^{\infty}a_{\lambda}\left[(\lambda+k)\{2(\lambda+k-1)-1\}+1\right]x^{\lambda+k}
\end{equation*}
\begin{equation*}
+\sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k+2} =0
\end{equation*}
or,
\begin{equation}
\sum\limits_{\lambda =0}^{\infty}a_{\lambda}\left[(\lambda+k)(2\lambda+2k-3)+1\right]x^{\lambda+k}+\sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k+2} =0 \tag{4.4.3}
\end{equation}
Equating the coefficient of lowest power of \(x\) (i.e., the coefficient of \(x^{k}\)) to zero. We get the indicial equation.
\begin{equation*}
a_{o}\left[k(2k-3)+1\right] =0
\end{equation*}
or,
\begin{equation*}
2k^{2}-3k+1 =0 \quad \text{Since,}\quad a_{o}\neq 0.
\end{equation*}
or,
\begin{equation*}
2k^{2}-2k-k+1 =0
\end{equation*}
or,
\begin{equation*}
(2k-1)(k-1)=0 \hspace{1cm} \Rightarrow k=1,\frac{1}{2}
\end{equation*}
and the difference of roots is not an integer.
Equating the coefficient of next higher power of \(x\) (i.e., \(x^{k+1}\)) to zero, we get
\begin{equation*}
a_{1}\left[(k+1)(2k+2-3)+1\right] =0
\end{equation*}
or,
\begin{equation*}
a_{1}\left[(k+1)(2k-1)+1\right] =0
\end{equation*}
or,
\begin{equation*}
a_{1}\left[2k^{2}-k+2k-1+1\right] =0
\end{equation*}
or,
\begin{equation*}
a_{1}\left[2k^{2}+k\right] =0
\end{equation*}
or,
\begin{equation*}
a_{1}\left[2k+1\right]k =0
\end{equation*}
for \(k=1\text{,}\)
\begin{equation*}
(2k+1)k \neq =0, \quad \text{hence,} \quad a_{1}=0
\end{equation*}
for \(k=\frac{1}{2}\)
\begin{equation*}
(2k+1)k \neq =0, \quad \text{hence,} \quad a_{1}=0
\end{equation*}
i.e., the series solution does not contain the coefficient \(a_{1}\text{.}\)
Again, equating the coefficient of general power of \(x\) (i.e., \(x^{\lambda+k+2}\)) to zero, we get the recursion relation.
\begin{equation*}
\sum a_{\lambda+2}\left[(\lambda+k+2)\{2k+2(\lambda+2)-3\}+1\right]+\sum a_{\lambda}=0
\end{equation*}
\begin{equation}
\therefore a_{\lambda+2} =-\frac{a_{\lambda}}{\left[(k+\lambda+2)\{2k+2\lambda+1\}+1\right]} \tag{4.4.4}
\end{equation}
for \(k=1\)
\begin{equation*}
a_{\lambda+2} =-\frac{a_{\lambda}}{\left[(\lambda+3)(2\lambda+3)+1\right]}
\end{equation*}
[from equation (4.4.4)] or,
\begin{equation*}
a_{2}=-\frac{a_{o}}{(3\cdot 3+1)}=-\frac{a_{o}}{10};\quad a_{4}=-\frac{a_{2}}{(5\cdot 7+1)}=\frac{a_{o}}{36\cdot 10}=\frac{a_{o}}{360};
\end{equation*}
and so on. Or,
\begin{equation*}
y_{1}=x\left[a_{o}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots\right]
\end{equation*}
\begin{equation}
\therefore y_{1}= a_{o}x\left[1-\frac{x^{2}}{10}+\frac{x^{4}}{360}-\cdots\right]\tag{4.4.5}
\end{equation}
for \(k=\frac{1}{2}\text{,}\)
\begin{equation*}
a_{\lambda+2} =-\frac{a_{\lambda}}{\left[(\lambda+\frac{5}{2})(2\lambda+2)+1\right]}
\end{equation*}
[from eqn. (4.4.4)] or,
\begin{equation*}
a_{2}=-\frac{a_{o}}{(\frac{5}{2}\cdot 2+1)}=-\frac{a_{o}}{6};
\end{equation*}
\begin{equation*}
a_{4}=-\frac{a_{2}}{(\frac{9}{2}\cdot 6+1)}=\frac{a_{o}}{6\cdot 28}=\frac{a_{o}}{168};
\end{equation*}
and so on. From eqn. (4.4.2), we have
\begin{equation*}
y_{2}=x^{\frac{1}{2}}\left[a_{o}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots\right]
\end{equation*}
\begin{equation}
\therefore, y_{2}= a_{o}x^{\frac{1}{2}}\left[1-\frac{x^{2}}{6}+\frac{x^{4}}{168}-\cdots\right] \tag{4.4.6}
\end{equation}
The complete solution of eqn. (4.4.1) is
\begin{equation*}
y=C_{1}y_{1}+C_{2}y_{2}
\end{equation*}
\begin{equation*}
=Ax\left[1-\frac{x^{2}}{10}+\frac{x^{4}}{360}-\cdots\right]+Bx^{\frac{1}{2}}\left[1-\frac{x^{2}}{6}+\frac{x^{4}}{168}-\cdots\right]
\end{equation*}
