What is the solution for xydx+sqrt(1+x^2)dy=0 ?

The solution for xydx+sqrt(1+x^2)dy=0 is y=e^{-sqrt(1+x^2)+c_{1}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xydx+sqrt(1+x^2)dy=0

x y d x + 1 + x^{2} d y = 0

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

Solution steps

x y d x + 1 + x^{2} d y = 0

Solve separable ODE:

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

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

y^{^{'}} + y cos (x) = 3 cos (x)

\frac{d y}{d x} + \frac{e ^{y^{2}}}{x ^{2} y} = 0, y (1) = 0

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

\frac{d x}{d t} - x^{3} = x

y^{^{'}} + 20 x y = 0

What is the solution for xydx+sqrt(1+x^2)dy=0 ?
The solution for xydx+sqrt(1+x^2)dy=0 is y=e^{-sqrt(1+x^2)+c_{1}}