What is the solution for y^'=3(y-1)^{2/3},y(0)=1 ?

The solution for y^'=3(y-1)^{2/3},y(0)=1 is y=t^3+1

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y^'=3(y-1)^{2/3},y(0)=1

y^{^{'}} = 3 (y - 1)^{\frac{2}{3}}, y (0) = 1

y = t^{3} + 1

Solution steps

y^{'} (t) = 3 (y - 1)^{\frac{2}{3}}

Solve separable ODE:

y = t^{3} + 1

y = t^{3} + 1

2 x y^{^{'}} = cos^{2} (y), y (0) = 2

\frac{d y}{d x} = \frac{2 x}{- 8 y - 6}

\frac{d y}{d x} = (y^{2} x)

2 y y^{^{'}} = 1, y (0) = x - 1

a (y) \frac{d y}{d t} = - a 2 g y

What is the solution for y^'=3(y-1)^{2/3},y(0)=1 ?
The solution for y^'=3(y-1)^{2/3},y(0)=1 is y=t^3+1