What is the solution for (dx}{dy}=\frac{y^3-3x)/y ?

The solution for (dx}{dy}=\frac{y^3-3x)/y is x=(y^3)/6+(c_{1})/(y^3)

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(dx}{dy}=\frac{y^3-3x)/y

\frac{d x}{d y} = \frac{y ^{3} - 3 x}{y}

x = \frac{y ^{3}}{6} + \frac{c _{1}}{y ^{3}}

Solution steps

\frac{d x}{d y} = \frac{y ^{3} - 3 x}{y}

Solve linear ODE:

x = \frac{y ^{3}}{6} + \frac{c _{1}}{y ^{3}}

x = \frac{y ^{3}}{6} + \frac{c _{1}}{y ^{3}}

n = 1 \sum \infty \frac{4}{2 n}

\frac{d ^{3}}{d x ^{3}} (\frac{6}{x ^{2}})

\int (y^{2} cos (x) - 3 x^{2} y - 2 x) d x

\frac{\partial}{\partial z} (ln (x y))

x^{3} - 4 x^{2}, 2 x^{2} - 8 x

What is the solution for (dx}{dy}=\frac{y^3-3x)/y ?
The solution for (dx}{dy}=\frac{y^3-3x)/y is x=(y^3)/6+(c_{1})/(y^3)