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

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

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y^'=y^3

y^{^{'}} = y^{3}

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

Solution steps

y^{'} (t) = y^{3}

Solve separable ODE:

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

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

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