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  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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 antiderivative is for some constant, C. Quote:
 

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