Making a substitution in a PDE - how did the author get to this result? 
How did the author get from $k_t = k^2 k_{\theta \theta} + k^3$ to that form I outlined? I've tried computing $k_{t}$ and $k_{\theta \theta}$ using $k(\theta, t)^2 = A(\theta) + B(t)$ and substituting that in the PDE, but it was all a mess and not close to what the author got. Also, why is it that $A''(\theta) + 4A(\theta)$ is a constant?
link to where I got this from. page 40
 A: Differentiating $k(\theta,t)^2=A(\theta)+B(t)$ twice with respect to $\theta$ and once with respect to $t$ yields
\begin{eqnarray*}
2kk_\theta&=&A'\;,\\
2k_\theta^2+2kk_{\theta\theta}&=&A''\;,\\
2kk_t&=&B'\;.
\end{eqnarray*}
Solve the first equation for $k_\theta$, substitute the result into the second equation, solve that for $k_{\theta\theta}$, solve the third equation for $k_t$ and substitute everything into the differential equation, multiply through by $k$, replace $k^2$ by $A+B$ everywhere, and you (should) get the desired result.
$A''+4A$ must be constant (with respect to $\theta$) because $(A''+4A)B$ is the only term that mixes $\theta$ and $t$. Consider the differential equation at fixed $\theta$ for two different times $t_1$ and $t_2$, and subtract the two resulting equations. The terms that depend only on $\theta$ cancel, so the only remaining term that depends on $\theta$ is $(A''(\theta)+4A(\theta))(B(t_2)-B(t_1))$. It follows that unless $B$ is constant (in which case the second factor would always be zero), the first factor cannot depend on $\theta$. 
