Solve the following differential equation: `(x^2-y^2)dx-2\ x y\ dy=0`

Explanation

`(x^2-y^2)dx-2\ x y\ dy=0`


`dy/dx=((x^2-y^2))/2xy`...(1)


So as this is a homogeneous differential equation


Putting `y=ux => u+(xdu)/dx=dy/dx`


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