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

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

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Emery 1 week 2021-09-09T14:30:51+00:00 1 Answer 0

Answers ( )

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    2021-09-09T14:32:02+00:00

    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(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{y^2},\\\\\\(ii)~\int\dfrac{1}{y}dy=\log y.

    From equation (i), we have

    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.

    Integrating the above equation on both sides, we get

    \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..

    Thus, the required solution of the given differential equation is

    x+y\log y=Cy..

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