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

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

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

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

y = t t, y = - t t

Solution steps

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

Solve separable ODE:

y = t t, y = - t t

y = t t, y = - t t

x^{^{'}} = \frac{( x ^{2} + x )}{( 2 x + 1 )}

3 = \frac{π h ^{2}}{4} \frac{d h}{d t}

sin (x) \frac{d y}{d x} + 3 y = x^{3} y

e^{- 2 x + 4 y} d x + e^{3 x - 6 y} d y = 0

2 x y d x = (3 - x^{2}) d y

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