What is the solution for y^'=x^2 1/y ?

The solution for y^'=x^2 1/y is y=sqrt((2x^3)/3+c_{1)},y=-sqrt((2x^3)/3+c_{1)}

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y^'=x^2 1/y

y^{^{'}} = x^{2} \frac{1}{y}

y = \frac{2 x ^{3}}{3} + c_{1}, y = - \frac{2 x ^{3}}{3} + c_{1}

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{3}}{3} + c_{1}, y = - \frac{2 x ^{3}}{3} + c_{1}

y = \frac{2 x ^{3}}{3} + c_{1}, y = - \frac{2 x ^{3}}{3} + c_{1}

y = x^{2 x}

\frac{d}{d x} (2 + 2 cos (x))

y = - x^{3} + 2, y = x - 3, x = - 1, x = 1

\int \frac{x}{y} d y

\frac{\partial}{\partial x} (ln (1 + x^{2} + y))

What is the solution for y^'=x^2 1/y ?
The solution for y^'=x^2 1/y is y=sqrt((2x^3)/3+c_{1)},y=-sqrt((2x^3)/3+c_{1)}