My Math Forum Help with differential equations proof!

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 May 28th, 2008, 01:50 PM #1 Newbie   Joined: May 2008 Posts: 24 Thanks: 0 Help with differential equations proof! Hi all, I would really appreciate some who can figure this to explain it to me. It's simply the proof for the test for exactness in differential equations. My teacher is dumb, I've asked her, but I dont think she even knows the subject. Anyway, here is a link to the proof that is in my book(because I can't figure this LaTeX thing out): http://www.teach.ustc.edu.cn/net_learn/ ... ma13np.htm The proof I'm looking at follows Theroem 13.1, near the top. I understand everything until the very end, where it says that dF/dY=N(X,Y) will hold so long as c'(y)=N(Xo,Y), but I've tried every thing and don't see this. ANY HELP IS MUCH APPRECIATED Eric
 May 28th, 2008, 11:53 PM #2 Newbie   Joined: May 2008 Posts: 19 Thanks: 0 Because df/dy=N, we must have (from equation (1)): df/dy=N(x,y)-N(x0,y)+c'(y)=N(x,y) By substracting N(x,y) from both sides, and adding N(x0,y), we see, that this is equivalent to c'(y)=N(x0,y) (equation). Then I think there is an error. It says that c'(y)=∫ (from y0 to y) N(x0,t) dt but obviously it should be c(y)=∫ (from y0 to y) N(x0,t) dt which is a "solution" to (equation). From this and equation (1) follows (2).
 May 29th, 2008, 03:34 AM #3 Newbie   Joined: May 2008 Posts: 24 Thanks: 0 tumppi, thanks for your reply! The thing is, we have to SHOW that df/dy=N(X,Y)) by showing that c'(y)=N(Xo,Y). We're given the fact that df/dx=M, and must shot that if dM//dy=dN/dx, then df/dy=N(X,Y), but it seems like the proof just keeps you hanging....
 May 29th, 2008, 04:05 AM #4 Newbie   Joined: May 2008 Posts: 19 Thanks: 0 But c is just the constant of integration, so we can choose it to be whatever we want to. (As a function of y only (not x) of course. Fortunately N(x0,y) is not a function of x )
 May 29th, 2008, 04:46 AM #5 Newbie   Joined: May 2008 Posts: 24 Thanks: 0 aha, now we're getting somewhere, lol. let me think about that for a second...

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