This is an ode from an old book, which I am stuck solving, since there is a term in front of $y''$ which does not have a power series. Here is the problem $$ x^{\frac{3}{2}} y'' + y = 0 $$ Expansion is at $x=0$. The book shows the solution using standard form, but I do not know how they obtained it. Here is a screen shot of the book with the solution it gives
I set $a=1$ for simplicity. Here is my attempt
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 $$ y=\sum_{n=0}^{\infty}a_{n}x^{n+r} $$ Then \begin{align*} 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*} Substituting the above back into the ode gives \begin{align} x^{\frac{3}{2}}\left( \sum_{n=0}^{\infty}\left( n+r\right) \left( n+r-1\right) a_{n}x^{n+r-2}\right) +\left( \sum_{n=0}^{\infty}a_{n} x^{n+r}\right) & =0\nonumber\\ \sum_{n=0}^{\infty}\left( n+r\right) \left( n+r-1\right) a_{n} x^{n+r-\frac{1}{2}}+\sum_{n=0}^{\infty}a_{n}x^{n+r} & =0\tag{1A} \end{align} To make all powers on $x$ the same, the above becomes \begin{equation} \sum_{n=0}^{\infty}\left( n+r\right) \left( n+r-1\right) a_{n} x^{n+r-\frac{1}{2}}+\sum_{n=\frac{1}{2}}^{\infty}a_{n-\frac{1}{2}} x^{n+r-\frac{1}{2}}=0\tag{1B} \end{equation} The indicial equation is obtained for $n=0$. From the above this gives $r(r-1)=0$. Hence the roots are $r_{1}=1$ and $r_{2}=0$. The roots differ by integer. Therefore we can construct two linearly independent solutions \begin{align*} y_{1}\left( x\right) & =x^{r_{1}}\sum_{n=0}^{\infty}a_{n}x^{n}\\ y_{2}\left( x\right) & =Cy_{1}\left( x\right) \ln\left( x\right) +x^{r_{2}}\left( \sum_{n=0}^{\infty}b_{n}x^{n}\right) \end{align*} Or \begin{align*} y_{1}\left( x\right) & =x\sum_{n=0}^{\infty}a_{n}x^{n}\\ y_{2}\left( x\right) & =Cy_{1}\left( x\right) \ln\left( x\right) +\sum_{n=0}^{\infty}b_{n}x^{n} \end{align*} Or \begin{align*} y_{1}\left( x\right) & =\sum_{n=0}^{\infty}a_{n}x^{n+1}\\ y_{2}\left( x\right) & =Cy_{1}\left( x\right) \ln\left( x\right) +\sum_{n=0}^{\infty}b_{n}x^{n} \end{align*} Where $C$ above can be zero. Let us find $y_{1}\left( x\right) $ first. From Eq(1B), where now $r=r_{1}=1$ $$ \sum_{n=0}^{\infty}\left( n+1\right) \left( n\right) a_{n}x^{n+\frac{1}% {2}}+\sum_{n=\frac{1}{2}}^{\infty}a_{n-\frac{1}{2}}x^{n+\frac{1}{2}}=0 $$ $n=0$ is skipped since used to find the roots. Let $a_{0}=1$. But how to find $a_{1},a_{2},\cdots$ since the index $n$ are not integers?
The whole problem starts because $x^\frac{3}{2}$ does not have power series expansion. This cause the index to become fractional. So I have no idea how to handle this case.
Maple claims there is no series solution to this ode
ode:=x^(3/2)*diff(y(x),x$2)+y(x)=0
infolevel[dsolve]:=3;
dsolve(ode,y(x),series)
dsolve/series/ordinary: vector Y of initial conditions at x0 = 0 [y(0) (D(y))(0)]
dsolve/series/ordinary: trying Newton iteration
dsolve/series/direct: trying direct subs
dsolve/series/froben: trying method of Frobenius
dsolve/SERIES: Warning: no solutions found
Then how did the book obtain one? How to handle such case?
As you have a fractional power in the equation, you have to assume that the series expansion also needs a fractional power of $x$ as the basis. So try $$ y(x)=\sum a_nx^{r+sn}\\ y'(x)=\sum (r+sn)a_nx^{r-1+sn} \\ y''(x)=\sum (r+sn)(r-1+sn)a_nx^{r-2+sn} \\ x^{3/2}y''(x)+y(x)=\sum (r+sn)(r-1+sn)a_nx^{r-1/2+sn}+\sum a_nx^{r+sn} $$ To get a solvable system, the second term of the first series has to be compatible with the first term of the second, which gives $s=\frac12$. The indicial equation for the lowest power is then $0=r(r-1)$, and for $r=1$ one gets the recursion for the coefficients $$ a_{n+1}=-\frac{a_n}{(n+1)/2(1+(n+1)/2)}=-\frac{4a_n}{(n+1)(n+3)}. $$ This directly gives the $A$ basis solution in the cited book.