What is the solution for y^'+4xy^3=0 ?

The solution for y^'+4xy^3=0 is y=sqrt(-1/(2(-2x^2+c_{1)))},y=-sqrt(-1/(2(-2x^2+c_{1)))}

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y^'+4xy^3=0

y^{^{'}} + 4 x y^{3} = 0

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

Solution steps

y^{'} (x) + 4 x y^{3} = 0

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

\frac{d}{d x} (sin (n x))

\frac{\partial}{\partial y} (ln (yz))

\int \frac{1}{x - y} d x

\int_{- 1}^{1} (13 - 12 u)^{- 1} d u

What is the solution for y^'+4xy^3=0 ?
The solution for y^'+4xy^3=0 is y=sqrt(-1/(2(-2x^2+c_{1)))},y=-sqrt(-1/(2(-2x^2+c_{1)))}