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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2xy^2+3x^2)dx+2x^2ydy=0

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{\partial}{\partial x} (\frac{1}{ln ( \frac{2}{x} )})

f (x) = 3 x^{2} - 1

f (x) = 12 x, (4, 24)

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 4

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