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

The solution for y^'=(2x^3)/(y^2),y(2)=1 is y=\sqrt[3]{(3x^4)/2-23}

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y^'=(2x^3)/(y^2),y(2)=1

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

y = 3 \frac{3 x ^{4}}{2} - 23

Solution steps

y^{'} (x) = \frac{2 x ^{3}}{y ^{2}}

Solve separable ODE:

y = 3 \frac{3 x ^{4}}{2} - 23

y = 3 \frac{3 x ^{4}}{2} - 23

\int_{0}^{5} π y^{3} d y

\frac{\partial}{\partial x} (\frac{200}{( x ^{2} + y ^{2} )})

\frac{2}{2 s + 1}

x \to 0 lim ((1 + 4 x) \frac{1}{x})

f (x) = x^{4}, at x = 4

What is the solution for y^'=(2x^3)/(y^2),y(2)=1 ?
The solution for y^'=(2x^3)/(y^2),y(2)=1 is y=\sqrt[3]{(3x^4)/2-23}