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} )}

y = sin^{3} (x^{2})

n = 1 \sum \infty (- \frac{7}{4})^{2}

\frac{cos ( x )}{2 + sin ( x )}

x^{3} ln (x + 1)

2 x csc (x) - x^{2} cot (x) csc (x)

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)))}