What is the solution for y^'=2xy^3 ?

The solution for y^'=2xy^3 is y=sqrt(-1/(2(x^2+c_{1)))},y=-sqrt(-1/(2(x^2+c_{1)))}

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y^'=2xy^3

y^{^{'}} = 2 x y^{3}

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

Solution steps

y^{'} (x) = 2 x y^{3}

Solve separable ODE:

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

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

\frac{d}{d t} (cos (t) + 1)

\frac{\partial}{\partial x} (A^{3} (\frac{x}{y})^{2} (1 - x))

\frac{d}{d x} (\frac{x ^{2} - 9}{x - 3})

\frac{d}{d x} (e^{x} (- x^{\frac{2}{3}} - \frac{4}{3} x^{- \frac{3}{2}}))

\int x^{2} (x^{3} + 2)^{4} d x

What is the solution for y^'=2xy^3 ?
The solution for y^'=2xy^3 is y=sqrt(-1/(2(x^2+c_{1)))},y=-sqrt(-1/(2(x^2+c_{1)))}