Q: How to separate this ODE? 
Find the general solution to the first-order differential equation $$(4-x^2)\frac{dy}{dx}+y=0$$

I'm painfully stuck on this ODE with no clear idea on how to approach solving it. The method I tried was wrong due to algebraic mishaps. Can someone please give me an idea on how to break this up so I can solve?

A: The equation is
$$\frac{dy}{dx} = y\cdot \frac{1}{x^2-4}$$
so it is of the type $$\frac{dy}{dx}=f(x)\cdot g(y)$$ where $g(y)=y$ and $f(x)=\frac{1}{x^2-4}$.

Now, apply the standard method of solving such equations, i.e. one that you probably learned in class and goes something like this. What do you get?

A: First, get the term without the derivative to the other side of the equation.
$$(4-x^2)\frac{dy}{dx} = -y\\
\frac{1}{y}\frac{dy}{dx} = \frac{1}{x^2-4}$$
Now you can separate the variables and use integration to get the final solution.

A: You can reformulate it by
$$ \frac{1}{y} \frac{{\rm d}y}{{\rm d}x}  = \frac{1}{x^2-4} $$
$$ \int \frac{1}{y}\,{\rm d}y = \int \frac{1}{x^2-4}\,{\rm d} x $$
$$ \ln y + C = \frac{1}{4} \ln \left( \frac{x-2}{x+2} \right) $$
