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

The solution for y^'=-ky^3 is y=sqrt(-1/(2(-kt+c_{1)))},y=-sqrt(-1/(2(-kt+c_{1)))}

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y^'=-ky^3

y^{^{'}} = - k y^{3}

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

Solution steps

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

Solve separable ODE:

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

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

\int (\frac{1}{x} + cot (x) - sec (x)) d x

e^{3 \cdot x} \cdot (2 \cdot sin (4 \cdot x))

\int \frac{1}{( 2 y + 5 ) ^{2}} d y

n = 1 \sum \infty (\frac{2}{3})^{n}

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{2 y}

What is the solution for y^'=-ky^3 ?
The solution for y^'=-ky^3 is y=sqrt(-1/(2(-kt+c_{1)))},y=-sqrt(-1/(2(-kt+c_{1)))}