I am trying to solve the PDE [math]\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2 \theta[/math]

I plugged [imath]u = R(r)T(\theta)[/imath] into the polar Laplacian formula and ended up with [imath]\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda[/imath]. My solutions to the Sturm-Louiville problems are [imath]R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}[/imath] and [imath]T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/imath] (I have ruled out the [imath]\lambda = 0[/imath] case). Thus I have [math]u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})[/math] Plugging in thee boundary condition and absorbing [imath]A + B[/imath] yields [math]\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/math] I tried forcing this into trig form as [imath]\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}[/imath] becomes via Euler's Formula [math]\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)[/math] How can I solve either of the blue equations for [imath]\lambda[/imath]?

I plugged [imath]u = R(r)T(\theta)[/imath] into the polar Laplacian formula and ended up with [imath]\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial T}{\partial t} = \lambda[/imath]. My solutions to the Sturm-Louiville problems are [imath]R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}[/imath] and [imath]T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/imath] (I have ruled out the [imath]\lambda = 0[/imath] case). Thus I have [math]u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta})[/math] Plugging in thee boundary condition and absorbing [imath]A + B[/imath] yields [math]\sin^2 \theta = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}[/math] I tried forcing this into trig form as [imath]\sin^2 \theta = Ce^{\sqrt{-\lambda}\theta i^2} + De^{-\sqrt{-\lambda}\theta i^2}[/imath] becomes via Euler's Formula [math]\sin^2 \theta = (C + D)\cos(\sqrt{-\lambda}\theta i) + (Ci - Di)\sin(\sqrt{-\lambda}\theta i)[/math] How can I solve either of the blue equations for [imath]\lambda[/imath]?

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