When solving one specific PDE to find $f(x,y)$, I came across the following equation through which I would find the eigenvalues $\lambda_l$: $$a\tanh^{-1}(\frac{\lambda}{b}) - \lambda = 0,$$ in which $a$ and $b$ are real constants. I found out graphically that when $a < b$, the values of $\lambda_l$ are real, but when $a > b$, they are purely imaginary, what makes the equation above become: $$a\tanh^{-1}(\frac{i\lambda}{b}) - i\lambda = 0,$$ which, reminding that $\tanh^{-1}(ix) = i\tan^{-1}(x)$, leads to: $$a\tan^{-1}(\frac{\lambda}{b}) - \lambda = 0,$$ For both real and imaginary cases, I have also found graphically 3 possible values for $\lambda_l: \lambda_1, \lambda_2$ and $\lambda_3$, so that $\lambda_1 = -\lambda_2$ and $\lambda_3 = 0$. The solution associated to any eigenvalue $\lambda_l$, $f_l(x,y)$, equals to: $$f_l(x,y) = \frac{a}{2}\frac{e^{{\lambda_l}x}}{b - {\lambda_l}y}.$$
Furthermore, it is also possible to see graphically that $∣\lambda_l∣ < ∣b∣$, and, due to the problem's physics, $y \in [-1, 1]$, so there is no possibility of a division by zero nor does the solution leads to infinite. Therefore, should I take into account this last eigenvalue $\lambda_3 = 0$ in my general solution $f(x,y)$, so that: $$f(x,y) = c_1\frac{a}{2}\frac{e^{{\lambda_1}x}}{b - {\lambda_1}y} + c_2\frac{a}{2}\frac{e^{{\lambda_2}x}}{b - {\lambda_2}y} + c_3\frac{a}{2b},$$ or should I discard the solution and the constant associated with $\lambda_3$?
