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} )}

\frac{1}{1 + cos ^{2} ( x )}

\int 4 x^{3} + 24 x - 1 d x

x \to 3 lim (\frac{3 - ∣ x ∣}{x + 3})

y^{^{''}} + 2 k y^{^{'}} + a k^{2} y = 0

\frac{s + 3}{s ^{2} + 4 s + 8}

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)))}