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First order linear non-homogeneous PDE

 
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Sep19-09, 11:29 PM   #1
 

First order linear non-homogeneous PDE


1. The problem statement, all variables and given/known data
Find the general solution to the PDE and solve the initial value problem:
y2 (ux) + x2 (uy) = 2 y2, initial condition u(x, y) = -2y on y3 = x3 - 2


2. Relevant equations/concepts
First order linear non-homogeneous PDEs


3. The attempt at a solution
I know that the general solution to the non-homogeneous PDE = a particular soltuion to it + the general solution to the assoicated homogenous PDE, so I first consider to assocatied homogeneous equation:
y2 (ux) + x2 (uy) = 0

The characteristic equation is dy/dx = x2/y2
And the solution to this ODE is y3 - x3 = C where C is an arbitrary constant.
So the general solution to the homogeneous equation is u = f(y3 - x3) where f is arbitrary differentiable function of one variable.
But then I am stuck. How can I find a particular solution to the original non-homogeneous PDE? Is there a systematic way to find one such solution?

Any help is appreciated! :)
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Sep20-09, 03:42 PM   #2
 
Can someone please help? I am still stuck on this problem... :(
Sep20-09, 05:16 PM   #3
 
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Quote by kingwinner View Post
Find the general solution to the PDE...
y2 (ux) + x2 (uy) = 2 y2
...

How can I find a particular solution to the original non-homogeneous PDE? Is there a systematic way to find one such solution?

Any help is appreciated! :)
I don't know about any systematic ways, but just looking at it, if you could find a u such that ux = 2 and uy = 0 that would do it. And that's easy.
Sep20-09, 10:56 PM   #4
 

First order linear non-homogeneous PDE


Quote by LCKurtz View Post
I don't know about any systematic ways, but just looking at it, if you could find a u such that ux = 2 and uy = 0 that would do it. And that's easy.
Then, I think u=2x will work as a particular solution.

But really, can anyone suggest a more general/systematic way to solve it that would work also for more complicated first order linear PDEs?

Thanks!!!
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