What is the solution for x^'=x^3(t) ?

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

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x^'=x^3(t)

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

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

Solution steps

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

Solve separable ODE:

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

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

x^{3} + 2 x + 1

f (x) = \frac{5}{x ^{2}}

\frac{x ^{2} - 1}{x}

csc (y) d x + sec^{3} (x) d y = 0

f (x) = x + 6, (10, 4)

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