Integrating dz=xdy+ydx

Heath

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Apr 21, 2003, 9:04:14 PM4/21/03

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Hi,
I can't remember how to integrate this equation:
dz=xdy+ydx
I tried using Linear 1st Order, Exact, and Homogeneous techniques, but
have not been able to get solution:
z=xy
Mathematica does solve it (if I solve xdy+ydx=0) , but I can't
remember how to solve dz=xdy+ydx by hand.
Thanks in advance for your help.
Heath

Charles Krug

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Apr 22, 2003, 9:08:19 AM4/22/03

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Sounds more like you're after, "Find a function whose gradient is xj +
yi"

If that's the case, you "simply" take the antiderivative of each term,
with the usual caveat that there are infinitely many functions F(x,y)
whose gradient is f(x,y).

Is that what you're after here?

GVK

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Apr 22, 2003, 9:27:33 AM4/22/03

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To solve equation
xdy+ydx=0
separate variable:
dy/y=-dx/x,
and integrate this eq.
ln(y)=-ln(x)+C,
then take exponential from both sides
y=A/x, where A=exp(C).
Your general solution is xy=A.

"Heath" <heath...@hotmail.com> wrote in message
news:f7e7d3d4.03042...@posting.google.com...

Martin Cohen

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Apr 22, 2003, 5:06:54 PM4/22/03

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How bout d(xy) = x dy + y dx = 0 so xy = constant?

Heath

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Apr 22, 2003, 6:03:35 PM4/22/03

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Hi Charles,
Another poster suggested the separable method:

"To solve equation
xdy+ydx=0
separate variable:
dy/y=-dx/x,
and integrate this eq.
ln(y)=-ln(x)+C,
then take exponential from both sides
y=A/x, where A=exp(C).
Your general solution is xy=A"

But this discounts the dz term. Is it possible to treat dz and xdy+ydx
separately in dz=xdy+ydx? Is the dz term the same as "A" in the above?

I'm not sure what you mean by "Find a function whose gradient is xj +
yi".
Thanks again,
Heath

cha...@pentek.com (Charles Krug) wrote in message news:<slrnbaafm7....@skinner.pentek.org>...

Heath

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Apr 22, 2003, 6:05:10 PM4/22/03

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Hi,
Thanks for your help. Is it valid to set dz=0 in dz=xdy+ydx then use
the separable method? If it is, then that's the solution I needed.
Thanks again,
Heath

"GVK" <kovalev...@qwest.net> wrote in message news:<Qmbpa.3$652....@news.uswest.net>...

GVK

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Apr 23, 2003, 8:15:26 AM4/23/03

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> > "Heath" <heath...@hotmail.com> wrote in message
> > news:f7e7d3d4.03042...@posting.google.com...
> >
> >>Hi,
> >>I can't remember how to integrate this equation:
> >>dz=xdy+ydx
> >>I tried using Linear 1st Order, Exact, and Homogeneous techniques, but
> >>have not been able to get solution:
> >>z=xy
> >>Mathematica does solve it (if I solve xdy+ydx=0) , but I can't
> >>remember how to solve dz=xdy+ydx by hand.
> >>Thanks in advance for your help.
> >>Heath
> >
> >
> >
> How bout d(xy) = x dy + y dx = 0 so xy = constant?

This is coincidence. How about xdy-ydx=0 or x^2dy+y^2dx=0?

Martin Cohen

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Apr 23, 2003, 2:23:08 PM4/23/03

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For x dy - y dx = 0, look at x/y.

For x^2dy+y^2dx=0, look at 1/x + 1/y.

No coincidence at all.

Martin Cohen

Narasimham G.L.

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Apr 23, 2003, 2:28:23 PM4/23/03

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heath...@hotmail.com (Heath) wrote in message news:<f7e7d3d4.03042...@posting.google.com>...

how to integrate this equation:

> dz=xdy+ydx > = d(xy) ; z=xy + constant