# Second Order D.E with non-constant (trig-functions) coefficients

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

$$-q''(z)+4 z q'(z)^3+\frac{3 q'(z)}{z}+3 q'(z)^2 \cot (q(z))+\frac{3 \cot (q(z))}{z^2} = 0$$

• Simplify the denominator first you dont need it – Isham Mar 15 '18 at 12:06
• @Isham Thanks, I had already done that and I updated the version in the original post. – Konstantinos Mar 15 '18 at 12:08
• @Isham Yes, indeed. My bad. I should have included the simplified version in the first place. Sorry. – Konstantinos Mar 15 '18 at 12:10
• I will ...but for Frobenius it's dead your equation is not linear and has trig functions... – Isham Mar 15 '18 at 12:39
• The change of variable is not valide you cant put $y=\arcsin(y)$ you can use $y=\arcsin(z)$...With the change of variable you made whats the value of z ? With $q(z)=arccot(q(z))$ ? – Isham Mar 15 '18 at 12:45