What is the solution for 3x(dy)/(dx)-2y=(x^3)/(y^2) ?

The solution for 3x(dy)/(dx)-2y=(x^3)/(y^2) is y=\sqrt[3]{x^3+c_{1}x^2}

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3x(dy)/(dx)-2y=(x^3)/(y^2)

3 x \frac{d y}{d x} - 2 y = \frac{x ^{3}}{y ^{2}}

Solution steps

3 x \frac{d y}{d x} - 2 y = \frac{x ^{3}}{y ^{2}}

Solve Bernoulli ODE:

\frac{d}{d x} (e^{- \frac{x ^{2}}{2} - \frac{y ^{2}}{2}})

e^{x}

\frac{\partial}{\partial x} (\frac{2 x - 2 z}{y + 2 z})

- 5 sin (3 t)

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

What is the solution for 3x(dy)/(dx)-2y=(x^3)/(y^2) ?
The solution for 3x(dy)/(dx)-2y=(x^3)/(y^2) is y=\sqrt[3]{x^3+c_{1}x^2}