What is the solution for (dy)/(dt)=y^3 ?

The solution for (dy)/(dt)=y^3 is y=sqrt(-1/(2(t+c_{1)))},y=-sqrt(-1/(2(t+c_{1)))}

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(dy)/(dt)=y^3

\frac{d y}{d t} = y^{3}

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

Solution steps

\frac{d y}{d 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} )}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 cos (3 t) - 20 sin (3 t)

f (x) = - x^{2} + 25, at x = 3

\int (1 + 2 x)^{- 1} d x

\int_{0}^{\infty} \frac{1}{x ^{2} + 4} d x

x^{4} e^{- x}

What is the solution for (dy)/(dt)=y^3 ?
The solution for (dy)/(dt)=y^3 is y=sqrt(-1/(2(t+c_{1)))},y=-sqrt(-1/(2(t+c_{1)))}