# partial differential equation, complete integral

• Nov 11th 2012, 07:26 PM
MAX09
partial differential equation, complete integral
This is my question.

Find the complete integral of the PDE

$\displaystyle \frac {\partial ^2 u}{\partial x^2} +2 \frac {\partial ^2 u}{\partial x \partial y}+ \frac{\partial ^2 u}{\partial y^2}$ = x ex+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...
• Nov 13th 2012, 08:04 AM
BobP
Re: partial differential equation, complete integral
Undetermined coefficients gets you there doesn't it ?
Assume that
$\displaystyle u(x,y) = f(x)e^{x+y},$
substitute, and deduce a second order linear ode for $\displaystyle f(x).$
• Nov 14th 2012, 06:18 AM
MAX09
Re: 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.
• Nov 14th 2012, 07:07 AM
BobP
Re: partial differential equation, complete integral
I think that your Auxiliary Equation is wrong, I think it should be $\displaystyle m^{2}+4m + 4 = 0$
having equal roots -2,-2.
For the particular integral, you can use the undetermined coefficients method, let $\displaystyle f(x)=Ax+B.$
• Nov 23rd 2012, 07:10 PM
MAX09
Re: 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,
• Nov 23rd 2012, 11:09 PM
JJacquelin
Re: 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).
• Nov 28th 2012, 01:29 AM
MAX09
Re: 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,

$\displaystyle \phi_{1}(y-x) +x \phi_{2}(y-x)$ ?

I guess it could also be

$\displaystyle \phi_{1}(x-y) +x \phi_{2}(x-y)$

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..
• Nov 28th 2012, 05:48 AM
JJacquelin
Re: 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.