Task:
Given a boundary value problem:
\begin{equation*}
\begin{cases}
L(x) = \ddot{x} + A(t)\dot{x} + B(t)x = 0
\\
\gamma(x) = x(t_0) = x_0
\\
\Gamma(x) = x(t_1) = x_1
\end{cases}
\end{equation*}
When, under such boundary conditions, does the solution of the problem exist, under the Jacobi condition, but not the strengthened Jacobi condition?
Definition:
We say that $\widehat{x}$ satisfies the Jacobi condition if the interval $(t_0, t_1)$ has no points conjugate to $t_0$, and the strengthened Jacobi condition if the interval $(t_0, t_1]$ there are no points conjugate to $t_0$.
Replacing the required function $$x(t) = y(t) + \frac{(t_1 - t)x_0 + (t - t_0)x_1}{t_1-t_0}$$
boundary conditions can be made homogeneous, and then we obtain the problem for the function $y(t)$ in the form
\begin{equation*}\label{eq:1.1} \begin{cases} L(y) = \ddot{y} + A(t)\dot{y} + B(t)y = f(t) \\ \gamma(y) = y(t_0) = 0 \\ \Gamma(y) = y(t_1) = 0 \end{cases} \end{equation*} where $$f(t) = -\frac{A(t)(x_1 - x_0) + (t_1 - t)x_0 + tx_1}{t_1-t_0}$$
Idea:
I wanted to use the theorem from differential equations:
Theorem. A necessary condition for the solvability of an inhomogeneous boundary value problem is the orthogonality of the right side of the equation $f(t)$ to the solution of the homogeneous problem $y(t)$.
$$\int_{t_0}^{t_1}f(t)y(t) = 0$$
Problem: From the statement of the problem follows, that the Jacobi condition is satisfied, which means that there is a conjugate point on the segment. Will this theorem be true if the Jacobi condition is satisfied?