What is the solution for (dv)/(dt)=v^{(2/3)} ?

The solution for (dv)/(dt)=v^{(2/3)} is v=(t^3}{27}+(c_{1}t^2)/9+(c_{1}^2t)/9+\frac{c_{1}^3)/(27)

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(dv)/(dt)=v^{(2/3)}

\frac{d v}{d t} = v^{(\frac{2}{3})}

v = \frac{t ^{3}}{27} + \frac{c _{1} t ^{2}}{9} + \frac{c _{1}^{2} t}{9} + \frac{c _{1}^{3}}{27}

Solution steps

\frac{d v}{d t} = v^{(\frac{2}{3})}

Solve separable ODE:

v = \frac{t ^{3}}{27} + \frac{c _{1} t ^{2}}{9} + \frac{c _{1}^{2} t}{9} + \frac{c _{1}^{3}}{27}

v = \frac{t ^{3}}{27} + \frac{c _{1} t ^{2}}{9} + \frac{c _{1}^{2} t}{9} + \frac{c _{1}^{3}}{27}

What is the solution for (dv)/(dt)=v^{(2/3)} ?
The solution for (dv)/(dt)=v^{(2/3)} is v=(t^3}{27}+(c_{1}t^2)/9+(c_{1}^2t)/9+\frac{c_{1}^3)/(27)