ydx+(y-x)dy=0Please be as thorough as possible when explaining this, I'm struggling very much trying to solve ODE's

Answer:  The required solution of the given differential equation is[tex]x+y\log y=Cy.[/tex]Step-by-step explanation:  We are given to solve the following ordinary differential equation :[tex]ydx+(y-x)dy=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]We will be using the following formulas for integration and differentiation :[tex](i)~d\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{y^2},\\\\\\(ii)~\int\dfrac{1}{y}dy=\log y.[/tex]From equation (i), we have[tex]ydx+(y-x)dy=0\\\\\Rightarrow ydx+ydy-xdy=0\\\\\\\Rightarrow \dfrac{ydx+ydy-xdy}{y^2}=\dfrac{0}{y^2}~~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by }y^2]\\\\\\\Rightarrow \dfrac{ydx-xdy}{y^2}+\dfrac{1}{y}dy=0\\\\\\\Rightarrow d\left(\dfrac{x}{y}\right)+d(\log y)=0.[/tex]Integrating the above equation on both sides, we get[tex]\int d\left(\dfrac{x}{y}\right)+\int d(\log y)=C~~~~~~~[\textup{where C is the constant of integration}]\\\\\\\Rightarrow \dfrac{x}{y}+\log y=C\\\\\Rightarrow x+y\log y=Cy.[/tex].Thus, the required solution of the given differential equation is[tex]x+y\log y=Cy.[/tex].