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?

  • $\begingroup$ Substitute $x^{1/2}=\alpha$ and calculate the method for the equation $$\alpha^4y''(\alpha^2)+(\alpha-K)y'(\alpha^2)=0.$$ $\endgroup$ – ty. Feb 17 '18 at 13:10
  • $\begingroup$ thanks @ty. In this case does my frobenius series look like $y(\alpha^2)=\sum_{n=0}^\infty a_n\alpha^{2n+s}$ $\endgroup$ – johnahh Feb 17 '18 at 14:02
  • $\begingroup$ No, it would be $(\alpha^2)^{n+s}=\alpha^{2(n+s)}. LutzL's substitution is easier though. $\endgroup$ – ty. Feb 17 '18 at 18:42

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 $$ y(x)=\sum a_n\sqrt{x}^{n+s}, y'(x)=\frac12\sum(n+s)\sqrt{x}^{n+s-2},... $$

  • $\begingroup$ I edited my equation, It wasn't supposed to have any $y'(x)$ terms. sorry. $\endgroup$ – johnahh Feb 17 '18 at 15:07

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