This is my question.

Find the complete integral of the PDE

= x e^{x+y}

involving arbitrary functions f1 and f2

A little nudge towards the right direction would be of great help. I'm sure I can take it from there...

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- November 11th 2012, 08:26 PMMAX09partial differential equation, complete integral
This is my question.

Find the complete integral of the PDE

= x e^{x+y}

involving arbitrary functions f1 and f2

A little nudge towards the right direction would be of great help. I'm sure I can take it from there... - November 13th 2012, 09:04 AMBobPRe: partial differential equation, complete integral
Undetermined coefficients gets you there doesn't it ?

Assume that

substitute, and deduce a second order linear ode for - November 14th 2012, 07:18 AMMAX09Re: partial differential equation, complete integral
Thank you BobP.

I could get as far as factoring the auxiliary equation and getting two equal roots as 1, and -1.

I guess the solution should be of the form (A+Bx)e^{-x}.

But i'm not able to get the next step to solve for the complete integral. - November 14th 2012, 08:07 AMBobPRe: partial differential equation, complete integral
I think that your Auxiliary Equation is wrong, I think it should be

having equal roots -2,-2.

For the particular integral, you can use the undetermined coefficients method, let - November 23rd 2012, 08:10 PMMAX09Re: partial differential equation, complete integral
BobP, can you post a link from where I can review my notes on finding the auxillary equation for such

partial differential equations? That'd be of great help as I can improvize and find the right set of solutions...

Thanks, - November 24th 2012, 12:09 AMJJacquelinRe: partial differential equation, complete integral
Hi !

The three main steps for solving the PDE are summarized in attachment (without the whole calculus, that you certainly can do by yourself). - November 28th 2012, 02:29 AMMAX09Re: partial differential equation, complete integral
Hi JJacquelin,

I solved the homogenous equation and got -1and -1 as the two roots.

Does this mean that the complementary function of the pde is,

?

I guess it could also be

As far as the particular integral is concerned, I'm a little at sea when i try following the image file you had attached with. Wish I was smart enough to get the gist of what you meant to explain.. - November 28th 2012, 06:48 AMJJacquelinRe: partial differential equation, complete integral
F(y-x) is solution of the homogeneous equation.

But since F is any function, F(y-x) is the same as G(x-y) with any function G , and the same as H(exp(y-x)) with any function H , and the same as... many others.

So you can chose any one of these functions containing (y-x). The function F(x-y)=F(-(y-x)) is one of them.

Now, to find a particular solution of the complete PDE, the method is already given in the preceeding post (part 2). Just apply it : bring back u=(ax+b)exp(x+y) into the PDE.