My Math Forum Solution of 2dx-e^(y-x)dy=0

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 August 8th, 2017, 10:35 AM #1 Newbie   Joined: Aug 2017 From: PK Posts: 1 Thanks: 0 Solution of 2dx-e^(y-x)dy=0 How can we solve this differential equation? Sent from my ALE-L21 using Tapatalk
 August 8th, 2017, 10:46 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,188 Thanks: 1648 Multiply by e^x, then integrate.
 August 10th, 2017, 03:22 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,244 Thanks: 887 Let u = y - x. Then y = u + x so dy = du + dx. The equation becomes 2dx + u(du + dx) = (2 + u)dx + udu = 0. So (2 + u)dx = -udu and then $\displaystyle dx= -\frac{u}{2+ u}du$. Now let v = 2 + u so the equation becomes $\displaystyle dx= -\frac{v- 2}{v}dv= -dv+ \frac{2}{v}dv$. Integrate and back substitute. Last edited by skipjack; August 10th, 2017 at 03:40 AM.
 August 10th, 2017, 03:51 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,188 Thanks: 1648 That's incorrect. The equation would become 2dx - e^u(du + dx) = (2 - e^u)dx - e^udu = 0, so (2 - e^u)dx = e^udu.
 August 10th, 2017, 01:40 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,244 Thanks: 887 Skipjack is correct- I completely lost the "e^" part! Fortunately, the correct result that he gives can be written as $\displaystyle dx= \frac{e^udu}{2- e^u}$. Letting $\displaystyle v= e^u$, $\displaystyle dv= e^u du$ so the equation becomes $\displaystyle dx= \frac{dv}{2- v}$. And then letting $w= 2- v$, $dw= -dv$ so $\displaystyle dx= -\frac{dw}{w}$, which is easy to integrate. Last edited by skipjack; August 10th, 2017 at 02:10 PM.

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