Second order differential equation using substitution

[Froskoy]

Homework Statement

$$\sin\theta\frac{d^2y}{d\theta^2}-\cos\theta\frac{dy}{d\theta}+2y\sin^3\theta=0$$

Homework Equations

Use the substitution $x=\cos\theta$

The Attempt at a Solution

I started off by listing:

$$x=\cos\theta\\ \frac{dx}{d\theta}=-\sin\theta\\ \frac{d^2x}{d\theta^2}=-\cos\theta\\$$

But don't know whether this helps, or where to go next. Could someone please give me a hint at how to approach this, I'd prefer not to have a full solution, but really am desperate for a starting point!

With very many thanks,

Froskoy.

 

[HallsofIvy]

Since you are differentiating with respect to $\theta$, you need to use the chain rule:
$$\frac{dy}{d\theta}= \frac{dy}{dx}\frac{dx}{d\theta}= -sin(\theta)\frac{dy}{dx}$$
Then
$$\frac{d^2y}{d\theta^2}$$
$$= \frac{d}{d\theta}(\frac{dy}{d\theta})$$
$$= \frac{d}{d\theta}(cos(\theta)\frac{dy}{dx}$$
$$= -sin(\theta)\frac{dy}{dx}+ cos(\theta)\frac{d}{d\theta}\frac{dy}{dx}$$
$$= -sin(\theta)\frac{dy}{dx}+ cos^2(\theta)\frac{d^2y}{dx^2}$$

 

[Froskoy]

Thanks very much! Have got it now!

 

[riversongs]

Hi! I was wondering why
d/dθ (dy/dθ ) does not =d/dθ (-sin(θ) dy/dx)?

Also, once you have the relevant expressions how do you solve the differential equation? Do you have to form the auxilliary equation? I tried to do that but it doesn't work out nicely.

Thanks!