What is the solution for y^'=8xy^3 ?

The solution for y^'=8xy^3 is y=sqrt(-1/(2(4x^2+c_{1)))},y=-sqrt(-1/(2(4x^2+c_{1)))}

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y^'=8xy^3

y^{^{'}} = 8 x y^{3}

y = - \frac{1}{2 ( 4 x ^{2} + c _{1} )}, y = - - \frac{1}{2 ( 4 x ^{2} + c _{1} )}

Solution steps

y^{'} (x) = 8 x y^{3}

Solve separable ODE:

y = - \frac{1}{2 ( 4 x ^{2} + c _{1} )}, y = - - \frac{1}{2 ( 4 x ^{2} + c _{1} )}

y = - \frac{1}{2 ( 4 x ^{2} + c _{1} )}, y = - - \frac{1}{2 ( 4 x ^{2} + c _{1} )}

\frac{d y}{d x} = 2 y

\frac{d y}{d x} = \frac{x sin ( x )}{y ^{2}}, y (0) = - 9

\frac{d y}{d x} 50 x^{\frac{1}{3}} y^{\frac{2}{3}} = 600

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

u^{^{'}} = e^{9 x - H (x)}

What is the solution for y^'=8xy^3 ?
The solution for y^'=8xy^3 is y=sqrt(-1/(2(4x^2+c_{1)))},y=-sqrt(-1/(2(4x^2+c_{1)))}