Question

Ydx-xdy = xydx Find differtial equation

Answer 1

hi mate here is your answer..

Answer 2

The required differential equation is $\log (\dfrac{x}{y}) =x+c.$

Step-by-step explanation:

We have,

$ydx-xdy = xydx$

Dividing both sides by xy, we get

$\dfrac{ydx-xdy}{xy} = \dfrac{xydx}{xy}$

⇒$\dfrac{dx}{x}-\dfrac{dy}{y} = dx$

Integrating both sides, we get

$\int {\dfrac{dx}{x}} \, -\int {\dfrac{dy}{y}} \, = \int {dx} \,$

⇒ $\log x-\log y=x+c$

[ ∵$\int {\dfrac{dx}{x}} =\log x$]

⇒  $\log (\dfrac{x}{y}) =x+c$

[ ∵ $\log x-\log y=\log (\dfrac{x}{y} )$]

Hence, the required differential equation is $\log (\dfrac{x}{y}) =x+c.$