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Q: Differential Equation : Coupled, Linear, First Order ( No Answer,   2 Comments )
Subject: Differential Equation : Coupled, Linear, First Order
Category: Science > Math
Asked by: johnpinckney-ga
List Price: $25.00
Posted: 25 May 2006 17:42 PDT
Expires: 24 Jun 2006 17:42 PDT
Question ID: 732451
Analytical solution for this problem:
f=f(x,y) g=g(x,y)
(d/dx)f + f = g
(d/dy)g + g = f
under conditions:
f(0,y)= 1
g(x,0)= 0
There is no answer at this time.

Subject: Re: Differential Equation : Coupled, Linear, First Order
From: sabertoothed123-ga on 30 May 2006 11:20 PDT
HI u know cant understand ur question ..............
Subject: Re: Differential Equation : Coupled, Linear, First Order
From: berkeleychocolate-ga on 06 Jun 2006 17:37 PDT
Here's some information suggesting that there is no solution with your
given boundary conditions. df/dx = g - f = - dg/dy. Take the
derivative of your first equation with respect to y and substitute for
dg/dy to get: d^2f/dxdy + df/dx + df/dy = 0. You probably knew that.
Now try a solution of the form: f(x,y) = u(x)*v(y). After some work
one gets that f(x,y) = A*exp(cx - cy/(c+1)) and g(x,y) = A(c+1)*exp(cx
- cy/(c+1)). Noticing that if f and g are solutions, then so are f + k
and g + k, we get solutions with three parameters: the above f + k and
the above g + k. None of these can fulfill your boundary conditions.

The pde in f is obviously linear homogeneous with solutions exp((cx -
cy/(c+1)) and 1. This makes it infinite dimensional since c can vary.
I guess that's what happens in pde's. There may be other solutions to
the equations, but I don't know what they would be.

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