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

The solution for y^'-2y^{1/3}=0,y(0)=0 is y=(8tsqrt(t))/(3sqrt(3)),y=-(8tsqrt(t))/(3sqrt(3))

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

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

y = \frac{8 t t}{3 3}, y = - \frac{8 t t}{3 3}

Solution steps

y^{'} (t) - 2 y^{\frac{1}{3}} = 0

Solve separable ODE:

y = \frac{8 t t}{3 3}, y = - \frac{8 t t}{3 3}

y = \frac{8 t t}{3 3}, y = - \frac{8 t t}{3 3}

y^{^{'}} - \frac{7 y}{5160} = 0.07

\frac{d y}{d x} = \frac{x cos ( x )}{y}, y (0) = - 2

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

\frac{d y}{d u} + 3 u^{2} y = 5 u^{2}

\frac{d Q}{d t} = \frac{50 - Q}{10}

What is the solution for y^'-2y^{1/3}=0,y(0)=0 ?
The solution for y^'-2y^{1/3}=0,y(0)=0 is y=(8tsqrt(t))/(3sqrt(3)),y=-(8tsqrt(t))/(3sqrt(3))