Theorem
The boundary value problem $$y''+p_1(x) y'+ p_2(x)=f(x)(1)\\ a_1 y(x_0)+a_2 y'(x_0)= b_1 y(x_1)+b_2 y'(x_1)=0 (2)$$
has a unique solution iff the corresponding homogeneous problem ( $f \equiv 0$) has only the trivial solution.
Proof
The solution of the equation is $y=y_{\mu}+ c_1 y_1+ c_2 y_2 (3)$
where $y_{\mu}$ is a solution of the non-homogeneous equation and $y_1, y_2$ are linearly independent solutions of the homogeneous problem.
Replacing $(3)$ at $(2)$ we get the following system as for $c_1, c_2$:
$$\begin{pmatrix} a_1 y_1(x_0)+a_2 y_1'(x_0) & a_1 y_2(x_0)+a_2 y_2'(x_0) \\ b_1 y_1(x_1)+b_2 y_1'(x_1) & b_1 y_2(x_1)+b_2 y_2'(x_1) \end{pmatrix} \begin{pmatrix} c_1\\ c_2 \end{pmatrix}=\begin{pmatrix} -a_1 y_{\mu}(x_0)-a_2 y_{\mu}'(x_0)\\ -b_1 y_{\mu}(x_1)-b_2 y_{\mu}'(x_1) \end{pmatrix}$$
which has a unique solution , iff the determinant of the matrix of coefficients is non-zero. But if this determinantis different from $0$, the corresponding homogeneous problem has only trivial solution, since we will have $c_1=c_2=0$.
Could you explain to me how from the fact that the matrix of coefficients is non-zero, we deduce that $c_1=c_2=0$