Solving an ODE with Legendre Polynomials

[CrosisBH]

TL;DR

Sol

From Griffiths E&M 4th edition. He went over solving a PDE using separation of variables. It got to this ODE
\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)= -l(l+1)\sin \theta \Theta
Griffths states that this ODE has the solution
\Theta = P_l(\cos\theta)
Where $$P_l = \frac{1}{2^l !} \frac{d^l}{dx^l} (x^2 - 1)^l $$ is a Legendre Polynomial. I was curious to see how this generalizes. I found the definition of the Legendre's Polynomials is
[source]

I have trouble seeing how this is the form of the ODE above. I've tried playing with it but I can't get it into the form where it makes sense that the Legendre Polynomials are the solution. I'm also curious of more examples of ODEs that can be manipulated into this form. Thank you!

 

[William Crawford]

Begin by deviding both side of
$$\frac{d}{d\theta}\bigg(\sin\theta\frac{d\Theta}{d\theta}\bigg) = - l(l+1)\sin\theta\,\Theta$$
by $\sin\theta$ to get
$$\frac{1}{\sin\theta}\frac{d}{d\theta}\bigg(\sin\theta\frac{d\Theta}{d\theta}\bigg) + l(l+1)\Theta = 0.$$
Now, introduce the new variable $x=-\cos\theta$ and thus
$$\frac{d}{d\theta} = \sin\theta\frac{d}{dx}.$$
Therefore
$$\frac{d}{dx}\bigg(\sin^2\theta\frac{d\Theta}{dx}\bigg) + l(l+1)\Theta = 0.$$
Lastly, notice that $x^2 = \cos^2\theta$ together with the trigonometric identity $\sin^2\theta = 1-\cos^2\theta$. Thus,
$$\frac{d}{dx}\bigg((1-x^2)\frac{d\Theta}{dx}\bigg) + l(l+1)\Theta = 0$$
which exactly is Legendre's differential equation.