2.4.1.5 Example 5. \(x^{\frac {3}{2}}y^{\prime \prime }+y=0\)

Solve \[ x^{\frac {3}{2}}y^{\prime \prime }+y=0 \] Since \(x=0\) is regular singular point, then Frobenius power series must be used. Let the solution be represented as Frobenius power series of the form \[ y=\sum _{n=0}^{\infty }a_{n}x^{\frac {n}{2}+r}\] Then \begin{align*} y^{\prime } & =\sum _{n=0}^{\infty }\left ( \frac {n}{2}+r\right ) a_{n}x^{\frac {n}{2}+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( \frac {n}{2}+r\right ) \left ( \frac {n}{2}+r-1\right ) a_{n}x^{\frac {n}{2}+r-2}\end{align*}

Substituting the above back into the ode gives \begin{align} x^{\frac {3}{2}}\left ( \sum _{n=0}^{\infty }\left ( \frac {n}{2}+r\right ) \left ( \frac {n}{2}+r-1\right ) a_{n}x^{\frac {n}{2}+r-2}\right ) +\left ( \sum _{n=0}^{\infty }a_{n}x^{\frac {n}{2}+r}\right ) & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \frac {n}{2}+r\right ) \left ( \frac {n}{2}+r-1\right ) a_{n}x^{\frac {n}{2}+r-\frac {1}{2}}+\sum _{n=0}^{\infty }a_{n}x^{\frac {n}{2}+r} & =0 \tag {1A}\end{align}

\(n=0\) gives the indicial equation

\begin{align*} \left ( r\right ) \left ( r-1\right ) a_{0}x^{r-\frac {1}{2}}+a_{0}x^{r} & =0\\ \left ( \left ( r\right ) \left ( r-1\right ) x^{r-\frac {1}{2}}+x^{r}\right ) a_{0} & =0\\ \left ( r\right ) \left ( r-1\right ) x^{r-\frac {1}{2}}+x^{r} & =0\\ \left ( \left ( r\right ) \left ( r-1\right ) x^{-\frac {1}{2}}+1\right ) x^{r} & =0\\ \left ( r\right ) \left ( r-1\right ) \frac {1}{\sqrt {x}}+1 & =0 \end{align*}

Not possible to obtain indicial equation in \(r\) only. How to handle this? Maple can’t solve this using series solution either.