Solving second order linear homogeneous differential equation

[the0]

Homework Statement

Find the set of functions from (-1,1)→ℝ which are solutions of:

(x^{2}-1)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-4y = 0

Homework Equations

The Attempt at a Solution

There is a hint which says to use the change of variable:
x=cos(θ)

doing this I get:

\frac{dx}{dθ} = -sin(θ)

\Rightarrow

dx = -sin(θ)dθ

\Rightarrow

(a): \frac{dy}{dx} = \frac{dy}{dθ}\frac{1}{-sin(θ)}

(b): \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)}

Can I do this?!?

If so, substituting everything in gives:

-sin^{2}(θ)\frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)} + cos(θ)\frac{dy}{dθ}\frac{1}{-sin(θ)} - 4y = 0

\Rightarrow

y'' + cot(θ)y' + 4y = 0

Now... I am not sure.
Have I made some mistake?
Or should I be able to solve this?
Could someone please point me in the right direction?
Thanks a lot!

 

[tiny-tim]

hi the0!

(a): \frac{dy}{dx} = \frac{dy}{dθ}\frac{1}{-sin(θ)}

(b): \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dθ^{2}}\frac{1}{sin^{2}(θ)}

no, your (b) doesn't include dy/dθ d/dx(-1/sinθ)