Solving (x²-1)y'' + xy' - 4y = 0 with substitution x = cos(θ)

[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)