What is the error I am making on getting general solutions for this Sturm-Liouville problem? Given this Sturm-Liouville problem:
$$X'' + \lambda X = 0$$
There are general solutions (Eigenfunctions) for three cases on $\lambda$:
$$\lambda > 0$$ Has the characteristic equation: $r^2+\lambda = 0$ with roots $r_1 = i \sqrt{\lambda}, r_2 = -i \sqrt{\lambda}$, inserting them into the derivative eigenfunction $e^{rx}$ results into: $$X(x) = c_1 e^{i \sqrt{\lambda}x} = c_1cos\sqrt{\lambda}x + c_2 sin \sqrt{\lambda}x$$
For the case:
$$\lambda < 0 $$
$$X(x) = c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x}$$
Finally, for case $\lambda = 0$ the characteristic equation is $r^2 = 0$, thus $r=0$ and:
$$X(x) = ce^{r x} = c$$
However the solution for this case is known to be:
$$X(x) = c_1 x + c_2$$ 
What is the error I am making? 
 A: $$ X^{''} + \lambda X\tag{1} $$
In addition, it is simpler to just call the second one hyperbolics. If $\lambda < 0$
$$  r = \pm \sqrt{ -\lambda}  \\ \lambda = -s   \tag{2}$$
$$ X^{''} = sX \tag{3}$$
$$ X(x) = c_{1}e^{\sqrt{s}x} + c_{2}e^{-\sqrt{s}x} \tag{4}$$
which gives us hyperbolics
$$ X(x) = c_{3}\cosh(\sqrt{s}x) +c_{4} \sinh(\sqrt{s}x) \tag{5} $$
in case $3$ we have to integrate
$$ X^{''} = 0\cdot X \\ 
X^{''} = 0  \tag{6} $$
integrate as usual
$$ X^{'} = \int 0 dx = c_{1} \tag{7}  $$
now integrate again
$$ \int X^{'}  = \int c_{1} dx \implies X(x) = c_{1}x + c_{2} \tag{8} $$
You don't need to use a characteristic equation. Or also that there are two roots that are $0$ so you need to make one have a polynomial power. You can't have two equal solutions. In the general problem is
$$ ay^{''} + by^{'} +cy =0 \tag{9}  $$
we have the characteristic equation
$$ ar^{2} +br + c= 0\tag{10}  $$
if we have double roots $r_{1} = r_{2} =r$
$$ y_{1}(t) =e^{r_{1}t} = e^{rt} ,y_{2}(t) =e^{r_{2}t} = e^{rt} \tag{11} $$
$$ r_{1,2} = \frac{-b \pm \sqrt{b^{2} -4ac }}{2a} \tag{12} $$
double roots come from
$$ r_{1,2} = \frac{-b}{2a} \tag{13} $$
$$ y_{1}(t) =  e^{\frac{-bt}{2a}} \tag{14}$$
to get a second solution you have multiply by $v(t)$ 
$$ y_{2}(t) = v(t)y_{1}(t) \tag{15} $$
you eventually figure out that $v(t) = t$
A: The special cases go away if you normalize the solutions at one endpoint, such as at $x=0$. For example, start by solving
$$
          X''+\lambda X = 0 \\
          X(0)=A,\;\; X'(0) = B.         
$$
The solutions then depend on $\lambda$ through a power series expansion. So special cases at, say $\lambda=0$, become limiting cases as $\lambda\rightarrow 0$, which allows you to ignore special cases during the process of finding the general solution. In this case, for $\lambda \ne 0$, the solutions are
$$
        X_\lambda(x)=A\cos(\sqrt{\lambda}x)+B\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.
$$
The solution when $\lambda=0$ is the limit of the above as $\lambda\rightarrow 0$:
$$
            X_{\lambda=0}(x)=A+Bx.
$$
But, if you do not normalize the endpoint conditions with constants, then the special case solutions are not necessarily limits of a family of solutions that you can find with arbitrary constants, which is something you have already discovered.
