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

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

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y^'=4xy^3

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

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}

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} )}

cos^{2} (x) sin (x) \frac{d y}{d x} + (cos^{3} (x)) y = 1

x \to - \infty lim (x^{2} - 2 x + 1 - x)

\int \frac{13}{( x ^{2} - 1 ) ^{2}} d x

\frac{\partial}{\partial y} (tan (x^{2} + y^{2}))

f (x) = \frac{1 - sin ( x )}{1 + sin ( x )}

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