Solve the following differential equation: `(1-x^2)dy+x y\ dx=x y^2dx`

Explanation

$$ \left(1-x^{2}\right) d y+x y d x=x y^{2} d x $$
Dividing by $$y^{2} d x$$, we get(and also divisible by $$\left(1-x^{2}\right)$$
$$ \frac{1}{y^{2}} \frac{d y}{d x}+\frac{x}{1-x^{2}} \frac{1}{y}=\frac{x}{1-x^{2}} $$
Now by putting $$\frac{-1}{y}=z$$
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