Let $a$ and $b$ two reals such as $ \ a<b$ and $I=\left[a,b\right]$. Let $q$ be a function continuous on $I$, we define the equation $$ y''-q(x)y=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(E_h\right) $$ We suppose that for $\left(\alpha,\beta\right)$ and $\left(\gamma,\delta\right)$ two couples of real not equal to $\left(0,0\right)$, the equation $\left(E_h\right)$ has two independant solutions $y_1$ and $y_2$ so that $$ \left\{\begin{matrix} \alpha y_1\left(a\right)+\beta y'_1\left(a\right)=0\\ \gamma y_2\left(b\right)+\delta y'_2\left(b\right)=0\\ \end{matrix}\right. $$ We now consider for $f$ continuous on $\ I$ the problem $(P)$ $$ \left(P\right)=\left\{\begin{matrix} y''-q(x)y=f\left(x\right)\\ \alpha y\left(a\right)+\beta y'\left(a\right)=0\\ \gamma y\left(b\right)+\delta y'\left(b\right)=0\\ \end{matrix}\right. $$
An article mentioned that $y$ is a solution of $\left(P\right)$ if and only if $$ y\left(x\right)=\int_{a}^{b}K\left(t,x\right)f\left(t\right)\text{d}t $$ where $$ K\left(t,x\right)=\frac{y_1\left(\text{min}\left(x,t\right)\right)y_2\left(\text{max}\left(x,t\right)\right)}{y'_1\left(x\right)y_2\left(x\right)-y_1\left(x\right)y'_2\left(x\right)}$$
But I think there is an error because it says that we also have $K\left(t,x\right)=K\left(x,t\right)$ which is not true here because of the denominator. Could someone explain to me if i'm wrong or correct this paper to make it clear to me ? I give you the link to the article but it is in French. My problem is at page 10.