I'm working on a project where I discuss using the metric tensor to compute the Laplacian on various Riemannian Manifolds, and how that can aid in solving certain Partial Differential Equations. In any case, I'm trying to solve a relatively simple PDE and I'm at a bit of an impasse. I fed Mathematica the following second-order PDE and I can't for the life of me figure out exactly how it reached this answer. I've tried the method of Frobenius, as well as power series solutions, and I just don't see it. In any case here is the mathematical code and result.
Input: DSolve[[Lambda]^2 R[r] + r (1 - 2 r^2) R'[r] + r^2 (1 - r^2) R''[r] == 0, R[r], r]
Output: {{R[r] -> C[1] Cosh[[Lambda] ArcTan[Sqrt[-1 + r^2]]] + I C[2] Sinh[[Lambda] ArcTan[Sqrt[-1 + r^2]]]}}
Latex: $$\text{Input: }\lambda^2R(r)+r(1-2r^2)R'(r)+r^2(1-r^2)R''(r)=0$$ $$\text{Output: }R(r)=C_1\cosh\left(\lambda\tan^{-1}\left(\sqrt{-1+r^2}\right)\right)+iC_2\sinh\left(\lambda\tan^{-1}\left(\sqrt{-1+r^2}\right)\right)$$
To be quite honest my computational skills are hardly the best, so I wouldn't be surprised if I'm making some trivial mistake, but I've wracked my brain and I still can't see it. Any help would be appreciated. Thank you.
For those interested the PDE is as follows: $$\begin{cases} \Delta_g u(r,\theta) = \left(\frac{1-2r^2}{r}\right)\frac{\partial u}{\partial r}+\left(1-r^2\right)\frac{\partial^2 u}{\partial r^2}+\left(\frac{1}{r^2}\right)\frac{\partial^2 u}{\partial \theta^2}=0\\ u(1,\theta)=10 \end{cases}$$ $\Delta_g$ is the Laplacian for the coordinate system defined $\left\{r\cos(\theta),r\sin(\theta),\sqrt{1-r^2}\right\}$
Let's define a new variable $x$ through the equation $$ r={\rm sech}\, x. \tag{1} $$ Then $$ \frac{d}{dr}=\left(\frac{dr}{dx}\right)^{-1}\frac{d}{dx}= -\frac{1}{\tanh x\,{\rm sech}\,x}\frac{d}{dx}, \tag{2} $$ and, after some algebra, the differential equation $$ \lambda^2R(r)+r(1-2r^2)R'(r)+r^2(1-r^2)R''(r)=0 \tag{3} $$ can be rewritten as $$ R''(x)+\lambda^2R(x)=0, \tag{4} $$ of which the general solution is $$ R(x)=A\cos(\lambda x)+B\sin(\lambda x). \tag{5} $$ From $(1)$ it follows that $$ 1-r^2=1-{\rm sech}^2\,x=\tanh^2x\Rightarrow x=\tanh^{-1}\left(\sqrt{1-r^2}\right), \tag{6} $$ so we may rewrite $(5)$ as $$ R(r)=A\cos\left(\lambda \tanh^{-1}\left(\sqrt{1-r^2}\right)\right) +B\sin\left(\lambda \tanh^{-1}\left(\sqrt{1-r^2}\right)\right). \tag{7} $$ To show that $(7)$ is equivalent to the solution given by Mathematica, we notice that $$ \tan^{-1}\left(\sqrt{-1+r^2}\right)=\tan^{-1}\left(i\sqrt{1-r^2}\right) =i\tanh^{-1}\left(\sqrt{1-r^2}\right) \tag{8} $$ and $$ \cosh(i\theta)=\cos(\theta),\qquad\sinh(i\theta)=i\sin(\theta). \tag{9} $$