1. ## u_{xx}+u_{yy}=0

Separation of Variables

$u_{xx}+u_{yy}=0$

$\displaystyle e^{rx+sy}=e^{rx}e^{sy}$

$u(x,y)=\varphi(x)\psi(y)$

$\varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$

$\displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

$\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

$\Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

How are those solutions obtained?

Also, I can't tell if the books has $e^{\pm y\sqrt{\lambda}}$ or $e^{\pm y^{\sqrt{\lambda}}}$.

2. Originally Posted by dwsmith
Separation of Variables

$u_{xx}+u_{yy}=0$

$\displaystyle e^{rx+sy}=e^{rx}e^{sy}$

$u(x,y)=\varphi(x)\psi(y)$

$\varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$

$\displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

$\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

$\Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

How are those solutions obtained?

Also, I can't tell if the books has $e^{\pm y\sqrt{\lambda}}$ or $e^{\pm y^{\sqrt{\lambda}}}$.
The two final DEs are second-order linear constant coefficient homogeneous. I suggest you read Differential Equations Tutorial to refresh your memory how to solve DEs of this type.

3. Originally Posted by Prove It
The two final DEs are second-order linear constant coefficient homogeneous.
Thanks