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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: 18,048 Thanks: 1395 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 $\frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy$. Given the above total differential, in order to find F you have to solve the two equations:
$\frac{\partial F}{\partial x}= (2 cos(x)- sin(x))e^{2x+ 3y}$ and
$\frac{\partial F}{\partial y}= 3e^{2x+ 3y}cos(x)$.

Integrate the first equation with respect to x, treating y as a constant. That would give $F(x,y)= e^{2x+ 3y}cos(x)$ 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 $F(x,y)= e^{2x+ 3y)cos(x)+ g(y)$ where g can be any (differentiable) function of y.

Differentiaing that F(x, y) with respect to y, $\frac{\partial F}{\partial y}= 3e^{2x+ 3y}cos(x)+ g'(y)$ and that must be equal to $2e^{2x+ 3y}cos(y)$. 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 $F(x,y)= e^{2x+ 3y}cos(x)+ C$ 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"!

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