$\frac{1}{\sin^2(t)}\frac{d}{dt}\sin^2(t)\frac{d}{dt}f=-k(k+2)f$

Question

I'm trying to solve the differential equation $$\frac{1}{\sin^2(t)}\frac{d}{dt}\left(\sin^2(t)\frac{d}{dt}f\right)=-k(k+2)f.$$ I know that (multiples of) $\frac{\sin((k+1)t)}{\sin(t)}$ are a solution but there have to be more solutions than this since the differential equation is of order two or am I wrong here? Thanks in advance.

Answer 1

Another linearly independent solution is $$ \frac{\cos((k+1)t)}{\sin(t)}. $$ You can find it using the method of reduction of order. The change $$ f(x)=\frac{\sin((k+1)t)}{\sin(t)}\,g(x) $$ will lead to a first order equation.

Another possibility is the change $$ g(t)=\frac{f(t)}{\sin t}, $$ which leads to a second order equation with constant coefficients.