What is the solution for y^'=y^{2/3} ?

The solution for y^'=y^{2/3} is y=(t^3}{27}+(c_{1}t^2)/9+(c_{1}^2t)/9+\frac{c_{1}^3)/(27)

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y^'=y^{2/3}

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

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

Solution steps

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

Solve separable ODE:

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

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

\int \frac{4}{- 3 x} d x

\frac{d}{d x} (I n^{2})

f (x) = \frac{x}{( x ^{2} + 15 x + 50 )}

1 - x, y = 2, - 1, 1

\frac{d}{d x} (x^{2} 3 - 4 x)

What is the solution for y^'=y^{2/3} ?
The solution for y^'=y^{2/3} is y=(t^3}{27}+(c_{1}t^2)/9+(c_{1}^2t)/9+\frac{c_{1}^3)/(27)