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May 12th, 2017, 04:51 PM   #1
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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.
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May 12th, 2017, 05:20 PM   #2
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Yes.
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May 15th, 2017, 03:01 AM   #3
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Quote:
Originally Posted by max233 View Post
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"!
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