What is the solution for xy^'+2y-2x^2y^2=0 ?

The solution for xy^'+2y-2x^2y^2=0 is y= 1/(x^2(-2ln(x)+c_{1))}

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xy^'+2y-2x^2y^2=0

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

Solution steps

x y^{'} (x) + 2 y - 2 x^{2} y^{2} = 0

Solve Bernoulli ODE:

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

What is the solution for xy^'+2y-2x^2y^2=0 ?
The solution for xy^'+2y-2x^2y^2=0 is y= 1/(x^2(-2ln(x)+c_{1))}