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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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}

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)