What is the solution for (dy)/(dx)=(yx+y)/(yx+x) ?

The solution for (dy)/(dx)=(yx+y)/(yx+x) is y=\W(x/(e^{-x-c_{1)}})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(yx+y)/(yx+x)

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

y = W (\frac{x}{e ^{- x - c_{1}}})

Solution steps

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

Solve separable ODE:

y = W (\frac{x}{e ^{- x - c_{1}}})

y = W (\frac{x}{e ^{- x - c_{1}}})

x d x - y d y = 0

\frac{\partial}{\partial z} (e^{y})

y = x^{4} - 9 x^{2}, y = x^{2} - 9

\int 6 (t + 1)^{\frac{1}{2}} d t

\frac{d y}{d x}, cos (x - y) = x e^{x}

What is the solution for (dy)/(dx)=(yx+y)/(yx+x) ?
The solution for (dy)/(dx)=(yx+y)/(yx+x) is y=\W(x/(e^{-x-c_{1)}})