Undetermined coefficients gets you there doesn't it ?
Assume that
substitute, and deduce a second order linear ode for
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...
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.
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 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.