I have the following equation
$$tx''-x=0$$
which I want to solve via power series.
I get it in "standard" form to apply Frobenious method. My initial equation becomes
$r(r-1)$.
After equating coefficents I get
$a_{n-1}=\dfrac{a_{n-2}}{n(n-1)} $ in front of $t^{n}$. For $n\ge 1$ and $a_{0}$ arbitrary. However according to the book it should be $a_{1} \sum_{k\ge 1}\dfrac{t^{k}}{k!(k-1)!}$.
I've done it three times. Can anyone get this right?
The duplicate does not accomodate my troubles, my problem was about renamning coefficents. The second possible duplicate is closer, but still not the same.
Your solution is correct, as is the one in the book. For the root $r=1$, with your notation, the solution is $$ t\sum_{n=0}^\infty a_n\,t^n=a_0\,t+a_1\,t^2+\dots. $$ What the book is doing is renaming $a_0\to a_1$, $a_1\to a_2$, ... and looking for solution of the form $$ \sum_{n=1}^\infty a_n\,t^n. $$