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} )}

\int 16 cos (x) d x

x^{'} = 0.0333333 x (30 - x)

x \to 2 lim (\frac{x ^{3} - 3 x ^{2} + 4}{3 x ^{3} - 18 x ^{2} + 36 x - 24})

\int_{4}^{\infty} 9 e^{- \frac{y}{2}} d y

k = 1 \sum \infty \frac{3 k ^{8}}{k !}

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)))}