Consider the following setup. We have a second order boundary value problem: $$\dfrac{d^2y}{dx^2}=F(x,y,dy/dx);\qquad y(x_0)=y_0,\quad y(x_f)=y_f.$$
A numerical approach is to almost first write as two first order ivps: $$\begin{aligned} \frac{dy}{dx}&=v,\qquad &y(x_0)=y_0 \\ \frac{dv}{dx}&=F(x,y,v),\qquad &v(x_0)=? \end{aligned}$$
Suppose we have some numerical method (some Runge-Kutta, mostly interested here in simple RK1, Euler's Method).
Now, we can take a guess at $v(x_0)$: for example we can shoot straight at $y(x_f)$ by using our numerical method to attack a system of IVPs by taking $v(x_0)=v_0$ where: $$v_0=\dfrac{y(x_f)-y(x_0)}{x_f-x_0}.$$ We implement our method and we get, as an approximation to $y(x_f)=y_f$, we get, say: $$y_f(v_0)\approx y_f.$$ Say the discrepancy is: $$\varepsilon(v_0)=y_f(v_0)-y_f.$$ Now we can plot $\varepsilon$ vs $v_0$ and where $\varepsilon$ is zero we have approximations to the solution that at least satisfy the second boundary condition.
When the original ode is suitably linear, it is the case for any two points $(v_a,y_a-y_f)$ and $(v_b,y_b-y_f)$ on this curve, we have $\varepsilon$ is zero at: $$v_a+\frac{y_f-y_a}{y_b-y_a}(v_a-v_b).$$ Because the curve in this case is a line. This formula can be understood on a more basic level if you start drawing pictures in the $x$-$y$ plane.
Is there a good, online and available reference that discusses this curve? What are the conditions on the ode on it being a straight line? What are the conditions on the ode that there is a unique root?
