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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

$X(x)=Ae^{-px}+Be^{px}$

$Y(y)=Ce^{-\sqrt{1-p^2}y}$+$De^{\sqrt{1-p^2}y}$

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

maxwell
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1 Answers1

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

Robert Israel
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  • 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 07 '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 07 '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 07 '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 07 '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 07 '13 at 17:41
  • For $1-p^2 < 0$, you want $\sqrt{p^2-1}$, not $\sqrt{1-p^2}$. – Robert Israel Jan 07 '13 at 18:32
  • Yeeah finally I got it correctly Sir. Thank you very much Sir for your kind help and effort. – maxwell Jan 08 '13 at 15:26