This is not a homework exercise. I am just making it clear.
I have the following second-order differential equation.
$$\frac{4 q'(z)^3 \sin ^3(q(z))}{z^2 \left(z^2 q'(z)^2+1\right)^{3/2}}+\frac{3 \sin ^2(q(z)) \cos (q(z))}{z^5 \left(z^2 q'(z)^2+1\right)^{3/2}}+\frac{3 q'(z) \sin ^3(q(z))}{z^4 \left(z^2 q'(z)^2+1\right)^{3/2}}+\frac{3 q'(z)^2 \sin ^2(q(z)) \cos (q(z))}{z^3 \left(z^2 q'(z)^2+1\right)^{3/2}}-\frac{q''(z) \sin ^3(q(z))}{z^3 \left(z^2 q'(z)^2+1\right)^{3/2}}=0$$
I know that the solution to the above is $ArcCos(m z)$, and this is something easily verifiable using a simple Mathematica code.
My question is the following:
How would I go about solving this by hand if I didn't have the solution?
From my undergraduate studies, I remember that $2^{nd}$ order D.E's with non-constant coefficients are used by applying the Frobenius Method; notes on the Frobenius method.
However I seem to be stuck with this, so any suggestions would be more than helpful.
Thanks in advance.
P.S: I am not asking anyone to solve it for me, just some recommendations would be nice. In particular, if I apply the Frobenius method, what do I do with the trig functions, as this is something I've never done before.
P.S: The simplified version is
$$-q''(z)+4 z q'(z)^3+\frac{3 q'(z)}{z}+3 q'(z)^2 \cot (q(z))+\frac{3 \cot (q(z))}{z^2} = 0$$