What is the solution for 3x^2y^2dx+(2y^3+4y^3)dy=0 ?

The solution for 3x^2y^2dx+(2y^3+4y^3)dy=0 is y=sqrt(-(x^3)/3+c_{1)},y=-sqrt(-(x^3)/3+c_{1)}

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3x^2y^2dx+(2y^3+4y^3)dy=0

3 x^{2} y^{2} d x + (2 y^{3} + 4 y^{3}) d y = 0

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

Solution steps

3 x^{2} y^{2} d x + (2 y^{3} + 4 y^{3}) d y = 0

Solve separable ODE:

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

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

\int_{0}^{3} e^{x^{2}} d x

x \to 0 lim (\frac{6}{x} - \frac{6}{x ^{2} + x})

\frac{\partial}{\partial x} (x^{7} y^{5} - x^{6} y^{4})

\frac{3}{5 s - 2}

\int \frac{2 x + 1}{x ^{3} - x ^{2}} d x

What is the solution for 3x^2y^2dx+(2y^3+4y^3)dy=0 ?
The solution for 3x^2y^2dx+(2y^3+4y^3)dy=0 is y=sqrt(-(x^3)/3+c_{1)},y=-sqrt(-(x^3)/3+c_{1)}