Originally Posted by

**dwsmith** Now I am solving the same equation using separation of variables. However, I have hit a snag.

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

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

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

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

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

$\displaystyle \displaystyle \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda=0$

$\displaystyle \displaystyle \psi''(y)-\psi'(y)-\lambda\psi(y)=0\Rightarrow n^2-n-\lambda=0$

The first equation is suppose to yield $\displaystyle s^2+\lambda$, and the second equation needs to be of the form $\displaystyle s^2-s+\lambda$ in order to obtain the same solutions from the exponential method.

I haven't come across this in the book but can I just set both equations equal to $\displaystyle -\lambda\mbox{?}$