We are given a differential equation, say of the form $$y''(z)+\frac Azy'(z)+\frac B{z^2}y(z)=0$$ This is a differential equation with regular singular points at $0$ and $\infty$. Then we find the general solution to this equation is something like $y(z)=a_1 z^{k_1}+a_2 z^{k_2}$ (given $A$ and $B$ do not satisfy particular conditions which would give repeated roots of the indicial equation rather than distinct $k_i$'s).
How can we now use a Mobius transform to find the general solution of a similar DE with regular singular points at generic point $x_1,x_2$ now? What is the general method? I can't find any similar questions on this site.
An idea I had was, of course we want a Mobius transform which maps $0,\infty$ to $x_1,x_2$. This would be $$f(z)=\frac{x_2z+x_1t}{z+t}$$ where $t$ is a free parameter. The inverse Mobius transform of this is $$f^{-1}(\tilde z)=\frac{\tilde z-x_1}{x_2-\tilde z}t$$Now we could let $z=f^{-1}(\tilde z)$ in the differential equation above, and this the solution above. In this case, whenever $(x_1,x_2)=(0,\infty)$, (by choosing $t=x_2$), we get that $f^{-1}$ is the identity map, and so we have recovered the original solution. If $x_i$ are different, then the general solution is $$y(z)=a_1 \left(f^{-1}(z)\right)^{k_1}+a_2\left(f^{-1}(z)\right)^{k_2}$$ Is that correct?
In particular I am unsure about the step where we "let $t=x_2$", and how valid that is - I'd expect that we wouldn't have needed to make a choice, and that this should work for any $t$. Did my derivation depend on the choice of $t$, and should it? These are my main concerns with my method.