Originally Posted by

**dwsmith** $\displaystyle u_{xyz}-xyzu=0$

$\displaystyle \varphi'(x)\psi'(y)\omega'(z)-xyz\varphi(x)\psi(y)\omega(z)=0$

$\displaystyle \displaystyle\frac{\varphi'(x)}{\varphi(x)}=\frac{ xyz\psi(y)\omega(z)}{\psi'(y)\omega'(z)}$

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

$\displaystyle \varphi(x)=\exp{(x\lambda)}$

$\displaystyle \lambda\psi'(y)\omega'(z)=xyz\psi(y)\omega(z)$

$\displaystyle \displaystyle\frac{\psi'(y)}{\psi(y)}=\frac{xyz\om ega(z)}{\lambda\omega'(z)}$

$\displaystyle \psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu$

$\displaystyle \psi(y)=\exp{(y\mu)}$

$\displaystyle \displaystyle\int\frac{\omega'(z)}{\omega(z)}=\int \frac{xyz}{\lambda\mu}dz\Rightarrow \ln{|\omega(z)|}=\frac{xyz^2}{2\lambda\mu}+c$

$\displaystyle \displaystyle\omega(z)=C_2\exp{\left(\frac{xyz^2}{ 2\lambda\mu}\right)}$

$\displaystyle \displaystyle u(x,y,z)=\varphi(x)\psi(y)\omega(z)=C\exp{\left(x\ lambda+y\mu+\frac{xyz^2}{2\lambda\mu}\right)}$

However, this solution doesn't check out.