Question

Ydx+(y-x)dy=0

Please be as thorough as possible when explaining this, I'm struggling very much trying to solve ODE's

Answer 1

Answer: The required solution of the given differential equation is

`x+y\log y=Cy.`

Step-by-step explanation: We are given to solve the following ordinary differential equation :

`ydx+(y-x)dy=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

We will be using the following formulas for integration and differentiation :

`(i)~d\left((x)/(y)\right)=(ydx-xdy)/(y^2),\\\\\\(ii)~\int(1)/(y)dy=\log y.`

From equation (i), we have

`ydx+(y-x)dy=0\\\\\Rightarrow ydx+ydy-xdy=0\\\\\\\Rightarrow (ydx+ydy-xdy)/(y^2)=(0)/(y^2)~~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by }y^2]\\\\\\\Rightarrow (ydx-xdy)/(y^2)+(1)/(y)dy=0\\\\\\\Rightarrow d\left((x)/(y)\right)+d(\log y)=0.`

Integrating the above equation on both sides, we get

`\int d\left((x)/(y)\right)+\int d(\log y)=C~~~~~~~[\textup{where C is the constant of integration}]\\\\\\\Rightarrow (x)/(y)+\log y=C\\\\\Rightarrow x+y\log y=Cy.`.

Thus, the required solution of the given differential equation is

`x+y\log y=Cy.`.