What is the solution for y^'=((1-x))/y ?

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

y^{^{'}} = \frac{( 1 - x )}{y}

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

Solution steps

y^{'} (x) = \frac{( 1 - x )}{y}

Solve separable ODE:

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

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

(u^{- 2} + u^{- 3}) (u^{5} + 5 u^{2})

\frac{d}{d x} (tan^{2} (x θ))

\int_{1}^{e^{5}} \frac{5}{x} d x

f (x) = 1 + 4 - x, at x = 2

x \to \infty lim ((x) 3)

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