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)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}

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)}