# Homework Help: Partial diff. eq. help

1. Feb 13, 2006

### schattenjaeger

UsubX = means partial of u with respect to x, for my halfasses notation...

so I have x^2*Usubx - Usuby + U = 0, I'm sposed to solve by a variable transformation thingy, and by seperation of variables. Doing the transformation I get U=const*e^(1/x) which is a solution, by my own checking. Yay

Now at this point I noted if that's the general solution(which I think that method finds)then there IS no seperated solution, and indeed despite my careful efforts whatever I get using SOV doesn't check out right. So assuming I'm right and didn't just muck it up, is this problem meant to teach us the potential failing of the SOV method? If so is there anyway to tell earlier it won't work besides doing it and checking?

And a semi-related question for clarification, if you use SOV(and get a solution that works) the solution is not guaranteed to be the most general solution, rather the most general seperated solution?

2. Feb 13, 2006

### schattenjaeger

Whoops, impressive what a minus sign will do, eh?

so I did get a more general solution, cool. So doing it by the variable transformation does NOT get you the most general solution? Or something?

3. Feb 14, 2006

### schattenjaeger

it's late but I see what I may've done, I didn't go through and solve the resulting ODEs setting the seperation paramater thing equal to 0, a negative const. and a positive const.

if upon doing that I see that I get some manner of trivial or no solutions for two cases, and one case that ends up as the e^(1/x) solution, I shall be very much reassured about how all this works. If not, do explain

Edit: But my answer I did get using SOV worked when I checked it. So whatever, I'll figure it out later

Last edited: Feb 14, 2006
4. Feb 14, 2006

### HallsofIvy

Of course, $U= Ce^{-\frac{1}{x}}$ is NOT the most general slolution. Remember that the most general solutions to partial differential equations have involve functions as the "constants" of integration. Since you haven't shown exactly what you have done, it is hard to say what you are doing. Separating U= X(x)Y(y) gives the two equations $X'= \frac{\lambdaX}{x^2}$ ( where $\lambda$ is an arbitrary constant), which has general solution $X(x)= Ce^{\frac{-\lambda}{x}}$and $Y'= \lambdaY+ 1$ which has solution $Y(x)= Ce^{(1+\lambda)y}$.

The general "separated" solution is of the form $U(x,y)= Ce^{\frac{-\lambda}{x}}e^{(1+\lambda)y}$.

What values $\lambda$ can take depends on the boundary conditions.

Of course, not every solution to the original equation is of that form- but every solution can be written as a (possibly infinite) linear combination of them.

Last edited by a moderator: Feb 14, 2006
5. Feb 14, 2006

### schattenjaeger

ok, thanks! I had a lightbulb over head moment in class today. I forgot that when you integrate partial diff. eqs that you don't get a constant, whether a function of the other variable, and in doing so it all works out happily ever after. And stuff