August 8th, 2017, 10:35 AM  #1 
Newbie Joined: Aug 2017 From: PK Posts: 1 Thanks: 0  Solution of 2dxe^(yx)dy=0
How can we solve this differential equation? Sent from my ALEL21 using Tapatalk 
August 8th, 2017, 10:46 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,689 Thanks: 1522 
Multiply by e^x, then integrate.

August 10th, 2017, 03:22 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807 
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: 18,689 Thanks: 1522 
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: 2,966 Thanks: 807 
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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2dxeyxdy0, solution 
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