What is the solution for y^'-8y^3=0 ?

The solution for y^'-8y^3=0 is y=sqrt(-1/(2(8t+c_{1)))},y=-sqrt(-1/(2(8t+c_{1)))}

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y^'-8y^3=0

y^{^{'}} - 8 y^{3} = 0

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

Solution steps

y^{'} (t) - 8 y^{3} = 0

Solve separable ODE:

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

\int \frac{1}{2 x - 1 - 4 2 x - 1} d x

\int x^{1.1} (\frac{1}{3 x} - 1) d x

x^{2} + y^{2} = 25

\int_{1}^{2} x (x - 1)^{9} d x

h \to 1 - lim (\frac{h - 2}{1 - h})

What is the solution for y^'-8y^3=0 ?
The solution for y^'-8y^3=0 is y=sqrt(-1/(2(8t+c_{1)))},y=-sqrt(-1/(2(8t+c_{1)))}