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.
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 $\lambda<0$ 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: $\lambda < 0$
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
$$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$$