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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(-x)/y

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

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

Solution steps

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

Solve separable ODE:

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

(x + 1) \frac{d y}{d x} (3 x + 1) y = x (x + 1) 1

(1 - x^{2}) y^{^{'}} - x y = 7 x y^{2}

\frac{d y}{d x} = e^{x - y}, y (0) = 0

\frac{d y}{d x} = 5 x y^{2}

\frac{d y}{d x} = y^{\frac{1}{3}}, y (0) = 0

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