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**ThePerfectHacker** $\displaystyle \underbrace{a^2 [u_{rr}+(1/r)u_r+(1/r^2)u_{\theta\theta}]}_{\mbox{Laplacian}} = u_t$

Let $\displaystyle u(r,\theta,t) = R(r)\Theta(\theta)T(t)$.

Thus,

$\displaystyle a^2[R''\Theta T + (1/r)R'\Theta T + (1/r^2)R\Theta ''T] = R\Theta T'$.

Thus,

$\displaystyle \frac{T'}{a^2T} = \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta}$

That means,

$\displaystyle \frac{T'}{a^2 T} = k \mbox{ and } \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta} = k$.

But the second equation can be rewritting as,

$\displaystyle \frac{r^2R''}{R} + \frac{rR'}{R} - kr^2 = - \frac{\Theta ''}{\Theta}$

Thus,

$\displaystyle \frac{r^2R''}{R}+\frac{rR'}{R} - kr^2 = c \mbox{ and } - \frac{\Theta '' }{\Theta} = c$.