Sturm–Liouville theory, linear differential equation We have Sturm–Liouville problem 
$$\left\{\begin{matrix}
{y}''(t)+4{y}'+3y=\lambda y\\ 
y(0)= y'(\ln(l))=0
\end{matrix}\right.
1<t<l$$
I've found roots of characteristic polynomial of differential equation
$$\tau = -2\pm\sqrt{1+\lambda }$$
So there are three three different solutions depending on $\lambda$.
$$1)\  \lambda = -1:\    y(t)= C_{1}e^{-2t}+C_{2}te^{-2t}$$
$$2)\ \lambda<-1: \ y(t)=C_{1}e^{-2t}\cos(\sqrt{-1-\lambda}t)+C_{2}e^{-2t}\sin(\sqrt{-1-\lambda}t)$$
$$3)\ \lambda>-1: \ y(t)=C_{1}e^{(-2+\sqrt{1+\lambda})t}+C_{2}e^{(-2-\sqrt{1+\lambda})t}$$
I don't have problmes with cases 1 and 2, but there is problem with case 3.
When we substitute the initial values, we get system of to equations:
$$\left\{\begin{matrix}
y(0)=C_{1}+C_{2} = 0 \rightarrow C_{1}=-C_{2} \\
y(\ln(l))=(-2+\sqrt{1+\lambda})C_{1}e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})C_{2}e^{(-2-\sqrt{1+\lambda})\ln(l)}=0\end{matrix}\right.$$
Then we get one equation:
$$(2-\sqrt{1+\lambda})C_{2}e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})C_{2}e^{(-2-\sqrt{1+\lambda})\ln(l)}=0$$
$$C_{2}\big((2-\sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})\ln(l)}\big)=0$$
And as I know, this equation have solution only when  $C_{2}=0 \rightarrow C_{1}=0$
So, the equation $$(2-\sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})\ln(l)}=0$$ Must not have solutions in real numbers, but I don't know how to prove that
 A: Let $f(t) = e^{-2t} \sinh{t}$.  We have $f'(t) = e^{-2t} (-2 \sinh{t} + \cosh{t})$.  Take
$$
t = \operatorname{arcosh}{\frac{2}{\sqrt{3}}}.
$$
We calculate 
$$
\sinh{t} = \sqrt{\cosh^2{t} - 1} = \sqrt{\frac{4}{3} - 1} = \frac{1}{\sqrt{3}}.
$$
So, if I am not mistaken, $f(t)$ is a solution of
$$
\left\{\begin{matrix}
{y}''+4{y}'+3y= 0
\\ 
y(0)= y'(\operatorname{arcosh}{\frac{2}{\sqrt{3}}})=0
\end{matrix}\right.
$$
with $\lambda = 0 > -1$.
A: Not quite. We want to solve
$$ (-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})t_0} - (-2 + \sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})t_0} = 0 $$
where $t_0 = \ln l$
Multiply through by $e^{2t_0}$ and let $\mu = \sqrt{1+\lambda}$ for brevity, we obtain
$$ (-2-\mu)e^{-t_0\mu} - (-2+\mu)e^{t_0\mu} = 0 $$
$$ 2(e^{t_0\mu} - e^{-t_0\mu}) - \mu (e^{t_0\mu} + e^{-t_0\mu}) = 0 $$
or
$$ \frac{\mu}{2} = \frac{e^{t_0\mu}-e^{-t_0\mu}}{e^{t_0\mu} + e^{-t_0\mu}} = \tanh (t_0\mu) $$
There is actually one solution if $t_0 > \frac12$ or $l > \sqrt{e}$. A closed form doesn't exist unfortunately, but you can see it graphically.
