What is the solution for y^'=(x-y)/(x+y) ?

The solution for y^'=(x-y)/(x+y) is y=(-2x+sqrt(8x^2+c_{1)})/2 ,y=(-2x-sqrt(8x^2+c_{1)})/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(x-y)/(x+y)

y^{^{'}} = \frac{x - y}{x + y}

y = \frac{- 2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{- 2 x - 8 x ^{2} + c _{1}}{2}

Solution steps

y^{'} (x) = \frac{x - y}{x + y}

Solve exact ODE:

y = \frac{- 2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{- 2 x - 8 x ^{2} + c _{1}}{2}

y = \frac{- 2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{- 2 x - 8 x ^{2} + c _{1}}{2}

\frac{d}{d x} (e^{2 x} 0.2)

x \to 0 lim (x cot (x))

10 x - 8 y x^{2} + 1 y^{^{'}} = 0

x \to 6 lim (6 (10) (- 5))

y = 0.0006 x^{0.837}

What is the solution for y^'=(x-y)/(x+y) ?
The solution for y^'=(x-y)/(x+y) is y=(-2x+sqrt(8x^2+c_{1)})/2 ,y=(-2x-sqrt(8x^2+c_{1)})/2