I am given the ODE: $$ (\sin{x})y''+2(\cos{x})y'-(\sin{x})y=0 $$ and asked to:
- Show $x=0$ is a regular, singular point
- Show both linearly independent solutions can be written as a power series about $x=0$
- Find the first two non-zero terms in each power series
I have no problem with the first part and was able to find that $p_0=2$ and $q_0=0$ (the values of the limits of $x*Q/P$ and $x^2*R/P$ as $x \rightarrow 0$ respectively). Using this I was able to find my indicial equation as: $$ r^2+r=0 \implies r=0,-1 $$ From here, I am not sure how to proceed. I believe my solutions should be of the form: $$ y_1(x)=\sum\limits_{n=0}^\infty a_nx^n \quad\quad\quad\text{and}\quad\quad\quad y_2(x)=\sum\limits_{n=0}^\infty a_nx^{n-1} $$ but I'm not entirely sure why. Is this where you have to check the continuity and differentiability of $P, Q, R$? Further, for the last part how do I find the coefficientss of the two solutions?
Any hints and general advice in solving ODEs using the Frobenius Method would be much appreciated!
EDIT: When trying to solve for the coefficients, should I expand $\sin{x}$ and $\cos{x}$ with Taylor Series?
Just for the sake of curiosity, I am going to explain @LutzLehmann's claim:
\begin{align*} \sin(x)y'' + 2\cos(x)y' - \sin(x)y = 0 & \Longleftrightarrow (\sin(x)y'' + \cos(x)y') + (\cos(x)y' - \sin(x)y) = 0\\\\ & \Longleftrightarrow (\sin(x)y'' + (\sin(x))'y') + (\cos(x)y' + (\cos(x))'y) = 0\\\\ & \Longleftrightarrow (\sin(x)y')' + (\cos(x)y)' = 0\\\\ & \Longleftrightarrow \sin(x)y' + \cos(x)y = c_{0}\\\\ & \Longleftrightarrow (\sin(x)y' + (\sin(x))'y)= c_{0}\\\\ & \Longleftrightarrow (\sin(x)y)' = c_{0}\\\\ & \Longleftrightarrow \sin(x)y = c_{0}x + c_{1}\\\\ & \Longleftrightarrow y(x) = \csc(x)(c_{0}x + c_{1}) \end{align*}
Hopefully this helps!