It was my exam question, and because of the question I had to dispute to my teacher about the method of solving this S.L. problem.
The equation:
$$y''+3y'+(1+\lambda^2 )y=0$$
with B.C. $$y(0)=y'(1)=0$$
$$-----------------------$$
My point is if the equation were given as $$y''+\lambda y=0$$ we know how to solve it first assume $\lambda>0$ then $<0$ then $=0$ etc.
However we do this because characteristic equation $r^2+\lambda=0$'s character changes according to neighborhood of $\lambda=0$(some kind of critical point.)
Therefore, if the equation were $y''+3y'+(1+\lambda^2 )y=0$, and just $\lambda>0$ is given,
then our charecteristic equation becomes $r^2+3r+1+\lambda^2=0$ then discriminant is $\triangle=5-4\lambda^2 $ so the solution of charecteristic equation changes according to $\lambda=\sqrt{5/4}$.
Should not we solve the Sturm-Liouville equation where $\lambda>\sqrt{5/4},<\sqrt{5/4}$ and $=\sqrt{5/4}$ eventhough $\lambda$ is given positive?
You really don't have to break things into cases if you normalize the functions at one or the other endpoint. So start by solving $$ y''+3y'+(1+\lambda^2)y = 0,\\ y(0)=0,\;y'(0)=1. $$ The general solution is determined by the equation $$ m^2 + 3m+1+\lambda^2 = 0 \\ (m+3/2)^2+(\lambda^2-5/4)=0 \\ m = -3/2\pm i\sqrt{\lambda^2-5/4}. $$
This gives solutions $y(x)=Ce^{3x/2}\sin(\sqrt{\lambda^2-5/4} x)$ where $C$ is chosen so that $$ 1=y'(0)=C\sqrt{\lambda^2-5/4} \\ \implies C = \frac{1}{\sqrt{\lambda^2-5/4}}. $$ So $$ y(x,\lambda)=\frac{e^{3x/2}\sin(\sqrt{\lambda^2-5/4}x)}{\sqrt{\lambda^2-5/4}}. $$ This works for all $\lambda$, even in the limit as $\lambda\rightarrow\pm\sqrt{5}/2$, where L'Hopital gives $$ y(x,\pm\sqrt{5}/2)=xe^{3x/2}. $$ If $\lambda^2-5/4 < 0$, then $\sin$ switches to $\sinh$ because $$ \sin(ix) = \frac{e^{i(ix)}-e^{-i(ix)}}{2i}= i\frac{e^{x}-e^{-x}}{2}=i\sinh(x). $$ Therefore for $\lambda^2-5/4 < 0$, $$ y(x,\lambda) = \frac{e^{3x/2}\sinh(\sqrt{5/4-\lambda^2}x)}{\sqrt{5/4-\lambda^2}}. $$ The special case at $\lambda^2=5/4$ gives the same limit as before.
The eigenvalue equation becomes a power series equation in $\lambda$: $$ y'(1)=0=\frac{3}{2}\frac{e^{3x/2}\sin(\sqrt{\lambda^2-5/4}x)}{\sqrt{\lambda^2-5/4}}+e^{3x/2}\cos(\sqrt{\lambda^2-5/4}x) $$ This can be written as a power series equation in $\lambda^2-5/4$.