I am trying to follow this derivation of the solution of the spherical lapace equation. This is probably very simple but where did $-m^2$ on the bottom of page one?
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Additionaly, what is happening between the bottom two lines. I know it says diff wrt r or $\theta$ but it doesnt look like the derivative has been taken.
http://www.physics.usu.edu/Wheeler/EM3600/Notes11SeparationOfVariablesSpherical.pdf
If $y(\varphi)=\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}$, then you effectively have an equation $dy/d\varphi = 0$, the solution is $y=\mathrm{const}=C$. However, if $C>0$, then the solution of equation $$ \frac{d^2\Phi}{d\varphi^2}=C\Phi >0,\qquad \Phi = C_1e^{\varphi \sqrt C}+C_2e^{-\varphi \sqrt C} $$ is non-periodic. However, we know that $\varphi$ is a cyclic variable, so $\Phi(\varphi)=\Phi(\varphi+2\pi)$. It means that $C\le0$. One of the way how to highlight it is to choose a constant that cannot be positive like $-m^2$.
In your second question, again let's denote functions to easily see what's going on: $$ \begin{aligned} f(r) &= \frac{1}{R} \frac{d}{d r}\left(r^{2} \frac{d R}{d r}\right) \\ g(\theta) &=\frac{1}{\Theta} \frac{1}{\sin \theta} \frac{d}{d \theta}\left(\sin \theta \frac{d \Theta}{d \theta}\right)-\frac{m^{2}}{\sin ^{2} \theta} \end{aligned} $$ You have an equation $f(r)+g(\theta)=0$, then you differentiate it by $r$, you will get $f'(r)=0$, the solution of this equation is $f(r)=\mathrm{const}=C_f$, by analogy the solution of $g'(\theta)=0$ is $g(\theta)=\mathrm{const}=C_g$. However, if you put those to the initial formula, you will get $C_f+C_g=0$, so clearly those constants are not independent. If you take $C_f=a$, then $C_g=-a$.