What is the solution for y^'=3\sqrt[3]{y^2} ?

The solution for y^'=3\sqrt[3]{y^2} is y=t^3+c_{1}t^2+(c_{1}^2t)/3+(c_{1}^3)/(27)

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y^'=3\sqrt[3]{y^2}

y^{^{'}} = 3 3 y^{2}

y = t^{3} + c_{1} t^{2} + \frac{c _{1}^{2} t}{3} + \frac{c _{1}^{3}}{27}

Solution steps

y^{'} (t) = 3 3 y^{2}

Solve separable ODE:

y = t^{3} + c_{1} t^{2} + \frac{c _{1}^{2} t}{3} + \frac{c _{1}^{3}}{27}

y = t^{3} + c_{1} t^{2} + \frac{c _{1}^{2} t}{3} + \frac{c _{1}^{3}}{27}

f (x) = x ln (x^{2})

(- 2, - 4), (- 3, - 3)

\frac{\partial}{\partial x} (e^{x y} y)

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

\frac{\partial}{\partial x} (e^{x} x)

What is the solution for y^'=3\sqrt[3]{y^2} ?
The solution for y^'=3\sqrt[3]{y^2} is y=t^3+c_{1}t^2+(c_{1}^2t)/3+(c_{1}^3)/(27)