What is the solution for 3x^2(1+y^2)dx-2ydy=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3x^2(1+y^2)dx-2ydy=0

3 x^{2} (1 + y^{2}) d x - 2 y d y = 0

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

Solution steps

3 x^{2} (1 + y^{2}) d x - 2 y d y = 0

Solve separable ODE:

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

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

y^{^{'}} = \frac{11 x ^{2} - 1}{11 + 3 y}

\frac{d y}{d x} = \frac{cos ( x ) y}{2}

y^{^{'}} = e^{- y} cos (t), y (\frac{π}{6}) = 1

\frac{d y}{d x} + \frac{3}{x} y = \frac{3}{x}

(6 \frac{d y}{d x}) - 36 y^{2} = 36

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