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} )}

\frac{d}{d x} (5 + tan^{2} (x))

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(3 + 4 x) + (5 y - 1) \frac{d y}{d x} = 0

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)))}