What is the solution for x^'=-x^3 ?

The solution for x^'=-x^3 is x=sqrt(-1/(2(-t+c_{1)))},x=-sqrt(-1/(2(-t+c_{1)))}

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x^'=-x^3

x^{^{'}} = - x^{3}

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

Solution steps

x^{'} (t) = - x^{3}

Solve separable ODE:

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

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

3 \frac{d x}{d t} + 2 x = 4, x = 0

u^{^{'}} = 2 e^{3 x - H (x)}

\frac{d y}{d x} - 8 \frac{d y}{d x} + 16 y = 0

y^{^{'}} = x (6 - y)

\frac{d y}{d x} = \frac{3 x + 3 x y ^{2}}{y x ^{2} + 2 y}

What is the solution for x^'=-x^3 ?
The solution for x^'=-x^3 is x=sqrt(-1/(2(-t+c_{1)))},x=-sqrt(-1/(2(-t+c_{1)))}