We can use the Frobenius method for an ODE like $$u'' + qu = 0$$ where the coefficients are functions if $q$ has a particular negative power series expansion. This is when we take infinity as a regular singular point. In this case, if the indicial equation gives us roots that integer difference, how do we write the solutions? In the case where we expand around 0, we have a term including $\ln(x)$ — what happens in the infinity case?
[Math] Frobenius method; expansion at infinity, what happens when difference of roots of indicial equation is integer
ordinary differential equations
Related Solutions
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.
If you look at http://math.creighton.edu/nielsen/DE_Fall_2010/Series%20Solutions/Series_Solutions_Beamer.pdf, they write the resulting recurrence for one of the solutions as $a_n F(n+r) = E$, where $F(r)$ is the indicial equation, and I'm writing $E$ to abbreviate a complicated expression which depends on a variety of things, including $n$. If $F(n+r)=0$ for some $n$ and $E \neq 0$, then this isn't solvable.
Best Answer
First you make the substitution $x\to\frac{1}{z}$, transform the ODE to the one in the variable $z$, and get a solution $y(z)$ which contains some $\ln(z)$. Then you just substitute back $z\to\frac{1}{x}$ and get terms with $\ln\frac{1}{x}$.