I am trying to solve an ODE using the Frobenius method. I understand the general process, but I do not understand how you compare coefficients when you have a $x^\frac{1}{2}$ term in the differential equatio. All my searches on google just go to non-integer differences in the indicial equation.
For example an equation such as
$x^2y''+(\sqrt{x}-K)y = 0 \\$
using the Frobenius series
$y = \sum_{n=0}^\infty a_n x^{n+s}$
Substituting this and its derivatives into the differential equation
$\sum_{n=0}^\infty a_n x^{n+s+2}+\sum_{n=0}^\infty a_n x^{n+s+1/2}-K\sum_{n=0}^\infty a_n x^{n+s} =0$
Normally I would make the first term start from $n=2$ or make $a_n\rightarrow a_{n-2}$ and do the same for the remaining terms to be able to compare powers of x.
How do you proceed when there is a non-integer power of x in the differential Equation?
Set $x=t^2$ and $u(t)=y(x)=y(t^2)$, then $u'(t)=2ty'(t^2)$ and $u''(t)=4t^2y''(t^2)+2y'(t^2)$ so that $$ t^2u''(t)=4t^4y''(t^2)+2t^2y'(t^2)=4(-(t-K)y(t^2))+2t^2y'(t^2) \\~\\ \implies t^2u''(t)-tu'(t)+4(t-K)u(t)=0 $$ so that now all exponents are integers and you can solve it with the standard Frobenius method.
Or start with $$ y(x)=\sum a_n\sqrt{x}^{n+s}, y'(x)=\frac12\sum(n+s)\sqrt{x}^{n+s-2},... $$