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} )}

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)))}