How to find eigenfunctions of a linear operator (follow-up question) This is a follow up to this question. My original question was answered by @oeanamen but for convenience here I state the follow up question completely.
I am interested in calculating characteristic values and eigenfunctions of $$K(x,t)=\max((1−x)t,(1−t)x),0<x<1,0<t<1$$ and find $λ_i$ and $y_i(x)$ such that
$$y_i(x)−λ_i\int_0^1K(x,t)y_i(t)dt=0.$$ 
After taking the second derivative of the above equation we find
$$y''=λy.$$
As oeanamen suggested in the comments to the original question a solution is $$A  \left(\sqrt{\lambda }
   \cosh \left(\sqrt{\lambda }
   x\right)-\sinh
   \left(\sqrt{\lambda }
   x\right)\right).$$
I realize that this is indeed a solution but I wonder if this is the only solution. I would also like to understand the details of calculation leading to the value of $\lambda$. 
My last question is what the corresponding eigenfunction is for $K(x,t)$ defined above. 
 A: As with the previous problem we can convert the integral equation into a differential equation by taking the second derivative of the integral equation with respect to $x$. 
We find 
$$y'' - \lambda y = 0.$$
We immediately throw away the solution for $\lambda =0$ ($y = A x + B$) since it implies $y = 0$ in the integral equation.
Thus, the solutions will be of the form
$$y = A \cosh\sqrt\lambda x + B \sinh\sqrt\lambda x.$$
In fact, by examining the integral equation and its first derivative evaluated at $x=0$ and $x=1$ we can convince ourselves that the solutions must satisfy Robin boundary conditions
$$\begin{eqnarray*}
y'(0) + y(0) &=& 0 \\
y'(1) - y(1) &=& 0.
\end{eqnarray*}$$
These boundary conditions make finding a closed form for the eigenvalues impossible. The solutions are peculiar. For $\lambda>0$ there is one eigenfunction. For $\lambda&lt0$ there is a tower of eigenfunctions. For large and negative $\lambda$ we will find approximate eigenvalues of the form $\lambda_n \approx -n^2\pi^2$.
The boundary conditions imply that $B = -A/\sqrt\lambda$ and that the eigenvalues satisfy the condition
$$\begin{equation}
\tanh\sqrt\lambda = \frac{2\sqrt\lambda}{1+\lambda}. \tag{1}
\end{equation}$$
Case I: $\lambda > 0$
There is one solution to equation (1) for $\lambda>0$. 
It must be found numerically. 
It is
$\lambda_0 \approx 2.38.$
The eigenfunction is 
$$y_0 = A(\sqrt\lambda_0 \cosh \sqrt\lambda_0 x - \sinh\sqrt\lambda_0 x).$$
Case II: $\lambda &lt 0$
Define $\mu = -\lambda$. 
The condition on the eigenvalues becomes
$$\begin{equation}
\tan\sqrt\mu = \frac{2\sqrt\mu}{1-\mu}. \tag{2}
\end{equation}$$
There is an infinite tower of countable solutions to equation (2). 
We find, for example, 
$$\mu_1 \approx 5.43 \approx \pi^2, \hspace{5ex}
\mu_2 \approx 35.4 \approx (2\pi)^2, \hspace{5ex}
\mu_3 \approx 84.8 \approx (3\pi)^2.$$
In the limit of large $\mu$, the right-hand side of (2) vanishes. 
Thus, for large $\mu$, $\sqrt\mu = n \pi$ will be an approximate solution, where $n\in\mathbb{N}$. 
(These are the positive zeros of the tangent function.)
That is, 
$\mu_n \approx n^2\pi^2$
for $n$ large. 
The eigenfunctions are
$$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$$
