Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi guys my name is Maxwell. This is my first time I asking question in this forum. I hope someone can help me this problem out :

Question says :

$ U_{xx} + U_{yy} = U $. Solve this PDE product solution using separation of variables.

What I do is

$ X''(x)Y(y) + X(x)Y''(y)=X(x)Y(y) $

$ X''(x)Y(y)=X(x)[Y(y)-Y''(y)] $

$ \frac{X''(x)}{X(x)} $ = $ \frac{Y(y)-Y''(y)}{Y(y)}$= k, where k is a constant

Then I made into 3 cases where $ k>0 $, $k<0$ and $k=0$

I already got the answer for $k<0$ and $k=0$ which my teacher say correct but for $ k>0 $ my teacher say wrong because he said for $ k>0 $ case, we need to divide into another 3 sub cases.

My $ k>0 [Let k=p^2 ] $, I got my answer



For this part could somone please solve it for me.Please dont say tips and hints. I need some work shown from you so that I can understand better. Please guys I really need help from you. This my first time in this forum. If someone could solve it, i will be really appreciate it. Thanks in advance

share|cite|improve this question
Look at this. – JohnD Jan 7 '13 at 17:10
up vote 2 down vote accepted

The three cases are $1-p^2 > 0$, $=0$, $< 0$.

share|cite|improve this answer
Why?? Could you show the final step please for the 3 sub cases. Please Im not good at PDE and I'm quite weak.I live in Malaysia and I dont have proper teacher to guide me. If you could show then I can analyse myself and try by myself later – maxwell Jan 7 '13 at 17:17
My X(x) is correct just the Y(y) only it is wrong.So help me to proceed the Y(y) only please – maxwell Jan 7 '13 at 17:24
This should be just like what you did for the $X$. Exponentials in one case, sine and cosine in another, $1$ and $y$ in the third. – Robert Israel Jan 7 '13 at 17:27
I really cant get it Sir. Really sorry Sir I'm totally blur about this. I still unable to get as what you said Sir. – maxwell Jan 7 '13 at 17:31
Is it like this now; for $1-p^2>0$, $Y(y)=Ce^{-\sqrt{1-p^2}y}$+$De^{\sqrt{1-p^2}y}$ and for $1-p^2<0$, $Y(y)=Ccos{\sqrt{1-p^2}y}$+$Dsin{\sqrt{1-p^2}y}$. Could you recheck my answer whether it is right or not Sir?? So that I can show to my teacher. – maxwell Jan 7 '13 at 17:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.