Originally Posted by

**dwsmith** The question now becomes what is the correct substitution for the exponential method so that it is equal to the separation of variables method (see below).

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

$\displaystyle \varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi''(y)\o mega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)=0$

$\displaystyle \displaystyle \frac{\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi' '(y)\omega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)}{\varphi(x)\psi(y)\omeg a(z)}=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}+\frac{\omega''(z)}{\omega(z)}-1=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda\Rightarrow m=\pm\sqrt{\lambda}$

$\displaystyle \displaystyle \lambda=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \displaystyle \frac{\psi''(y)}{\psi(y)}=\mu\Rightarrow \psi''(y)-\mu\psi(y)=0\Rightarrow n^2-\mu\Rightarrow n=\pm\sqrt{\mu}$

$\displaystyle \displaystyle \mu=1-\lambda-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \displaystyle \omega''(z)-\omega(z)+\lambda\omega(z)+\mu\omega(z)=0\Rightarr ow a^2-1+\lambda+\mu\Rightarrow a=\pm\sqrt{1-\lambda-\mu}$

$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\pm\sqrt{\lambda})}*\e xp{(y\pm\sqrt{\mu})}*\exp{(z\pm\sqrt{1-\lambda-\mu})}$

I can't figure out the correct substitution for the exponential method to match the separations method.