Thread: partial 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...

Undetermined coefficients gets you there doesn't it ?
Assume that

substitute, and deduce a second order linear ode for

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.

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

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,

Hi !
The three main steps for solving the PDE are summarized in attachment (without the whole calculus, that you certainly can do by yourself).

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..

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.