In fact Frobenius method is just an extension from the power series method that you add an additional power that may not be an integer to each term in a power series or even add the log term for the assumptions of the solution form of the linear ODEs so that you can find all groups of the linearly independent solutions that in cases of cannot find all groups of the linearly independent solutions when using power series method.
You can force to use Frobenius method when you find that the linear ODEs can already find all groups of the linearly independent solutions when using power series method, however, you will find that you can already find all groups of the linearly independent solutions when the additional power is just taking an non-negative integer and no need to add the log term.
If you doubt that whether the linear ODEs can find all groups of the linearly independent solutions when using power series method or not, you can feel free to use Frobenius method instead. The only thing is you should write more steps.
For example you force to solve $y''+xy=0$ by Frobenius method:
Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}$
$\therefore\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+x\sum\limits_{n=0}^\infty a_nx^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+\sum\limits_{n=0}^\infty a_nx^{n+r+1}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+\sum\limits_{n=3}^\infty a_{n-3}x^{n+r-2}=0$
$r(r-1)a_0x^{r-2}+r(r+1)a_1x^{r-1}+(r+1)(r+2)a_2x^r+\sum\limits_{n=3}^\infty((n+r)(n+r-1)a_n+a_{n-3})x^{n+r-2}=0$
$\therefore r=1,0,-1,-2$
When we take $r=0$ ,
$2a_2+\sum\limits_{n=3}^\infty(n(n-1)a_n+a_{n-3})x^{n-2}=0$
As we can already find all groups of the linearly independent solutions from this relation when we take $r$ as non-negative integer, and when taking $r$ as non-negative integer, the assumptions of the solution form is as same as assuming the solution form as power series. This implies that we can solve $y''+xy=0$ by power series method.
Your first solution is a power series. Your second solution is the Frobenius solution which allows for $r$ non-integer. There are 4 cases to consider for the second solution. The way in which the second solution is found differs slightly if:
- you have distinct exponents which do not differ by an integer
- you have distinct exponents which do differ by an integer
- you have repeated exponents
- you have complex exponents
The simplest example illustrating these is the Cauchy-Euler problem $ax^2y''+bxy'+cy=0$ which is the quintessential example of a second order ODE with a regular singular point.
To answer your question, if you seek a solution at an ordinary point use the power series solution. If you face a regular singular point then invoke the method of Frobenius.
Best Answer
Your series expansion is taken about the point $x_{0}$, which may be an ordinary, regular singular, or irregular singular point.