Find the general solution to the following differential equation : \(\frac{dy}{dx}\) =  -4xy2 , y(0) = 1

$y = \frac{1}{2x^2} + 1 $ y = \(\frac{1}{2x^2c}\) y = \(\frac{C}{2x^2}\) y = \(\frac{1}{2x^2+1}\)

y = \(\frac{1}{2x^2+1}\)

dy = -4xy2 . dx and then integrate to get :  $y = \frac{1}{2x^2} + C $