# How to Solve PDE using techniques of Separation Variables in this question [SOLVED]

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

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Look at this. – JohnD Jan 7 '13 at 17:10

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