How do I solve the following differential equation (It's not seperable)? I'm trying to Solve the following equation: find the solution $y:(-1,1) \rightarrow \mathbb{R}$ of $y'=\dfrac{y}{1-x^2}+x$?
It is not separable and I have no other Tools to solve it.

Subtract $ \frac {y(x)}{1-x^2} $ from both sides :
$ \frac {dy(x)}{dx} + \frac {y(x)}{x^2-1} = - \frac {-x^3 + x}{x^2-1}$ 
Let $ m(x) = e^{\int \frac {1}{x^2-1} dx } = \frac {\sqrt{1-x}}{\sqrt{x+1}} $
Then multiplay both sides by m(x) and substitute : $ \frac {\sqrt{1-x}}{\sqrt{x+1}(x^2-1)} = \frac {d}{dx} \frac {\sqrt{-x+1}}{\sqrt{x+1}} $. Apply the reverse product rule : $ g \frac{df}{dx} + f \frac{dg}{dx} = \frac{d}{dx}(fg) $ to the left side of the equation above and integrate both sides :
$\int \frac {\sqrt{-x+1}y(x)}{\sqrt{x+1}} dx= \int - \frac {\sqrt{-x+1}(-x^3 + x)}{\sqrt{x+1}(x^2-1)}dx $ 
You will then get :
$ \frac {\sqrt{-x+1}y(x)}{\sqrt{x+1}} = - sin^1(\frac{\sqrt{x+1}}{\sqrt{2}}) + \frac{1}{2}(x-2)\sqrt{-x^2 +1} + c $
Divide both sides by m(x) that we defined above and you get :
$ y(x) = \frac{\sqrt{x+1}(- sin^1(\frac{\sqrt{x+1}}{\sqrt{2}}) + \frac{1}{2}(x-2)\sqrt{-x^2 +1} + c)}{\sqrt{-x+1}} $
This is the solution to your differential equation, I hope I didn't do any typing mistake while writing it down !
