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 May 12th, 2017, 04:51 PM #1 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 Total differential function If I have this total differential: $\displaystyle (2\cos(x)-\sin(x))e^{(2x+3y)}dx+3e^{(2x+3y)}\cos(x)dy$ I then find the differential: $\displaystyle e^{(2x+3y)}\cos(x)$ I now need to find the function subject to the condition (0,0)=2 Is this just a case of finding the constant C, such as: $\displaystyle 2=e^{(2x+3y)}\cos(x)+C$ With C=1 Therefore the function is: $\displaystyle e^{(2x+3y)}\cos(x)+1$ Last edited by skipjack; May 12th, 2017 at 05:13 PM. May 12th, 2017, 05:20 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Yes. May 15th, 2017, 03:01 AM   #3
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Quote:
 Originally Posted by max233 If I have this total differential: $\displaystyle (2\cos(x)-\sin(x))e^{(2x+3y)}dx+3e^{(2x+3y)}\cos(x)dy$ I then find the differential: $\displaystyle e^{(2x+3y)}\cos(x)$
Actually, the first expression was, as you said, the "differential". What you are finding here is one possible anti-derivative.

Given a function F(x,y), the "total differential" is . Given the above total differential, in order to find F you have to solve the two equations:
and
.

Integrate the first equation with respect to x, treating y as a constant. That would give plus the "constant of integration". But since we are treating y as a constant, that "constant of integration might be a function of y. That is, we have where g can be any (differentiable) function of y.

Differentiaing that F(x, y) with respect to y, and that must be equal to . That means that g'(y)= 0 and, since it is a function of y only, g(x) is a constant. The general anti-derivative is for some constant, C.

Quote:
 I now need to find the function subject to the condition (0,0)=2 Is this just a case of finding the constant C, such as: $\displaystyle 2=e^{(2x+3y)}\cos(x)+C$ With C=1 Therefore the function is: $\displaystyle e^{(2x+3y)}\cos(x)+1$
Yes, but you shouldn't have waited until you were matching the condition to write the "+ C"! Tags differential, function, total Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post max233 Calculus 6 May 6th, 2016 08:21 PM golomorf Differential Equations 0 December 1st, 2014 03:34 AM adagus Differential Equations 1 November 19th, 2014 06:27 AM piotrek Differential Equations 2 May 23rd, 2013 07:22 AM Taurai Mabhena Differential Equations 11 January 30th, 2012 02:55 PM

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