These exercises outline an approach to solving linear two-point boundary value problems known as the shooting method. Exercises $31-34$ apply this method to solve specific problems.
We assume that the linear two-point boundary value problem,
$$
\begin{aligned}
&y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t), \quad a<t<b \\
&a_{0} y(a)+a_{1} y^{\prime}(a)=\alpha \\
&b_{0} y(b)+b_{1} y^{\prime}(b)=\beta,
\end{aligned}
$$
has a unique solution. As earlier, we assume that $\left|a_{0}\right|+\left|a_{1}\right|>0$ and $\left|b_{0}\right|+\left|b_{1}\right|>0$.
Let $y_{1}(t)$ and $y_{2}(t)$ denote solutions of the following two initial value problems:
$$
\begin{aligned}
&y_{1}^{\prime \prime}+p(t) y_{1}^{\prime}+q(t) y_{1}=g(t) \\
&y_{1}(a)=\alpha c_{1}, \quad y_{1}^{\prime}(a)=-\alpha c_{0}
\end{aligned} \text { and } \begin{aligned}
&y_{2}^{\prime \prime}+p(t) y_{2}^{\prime}+q(t) y_{2}=0 \\
&y_{2}(a)=a_{1}, \quad y_{2}^{\prime}(a)=-a_{0}
\end{aligned}
$$
where $c_{0}$ and $c_{1}$ are any two constants satisfying $a_{0} c_{1}-a_{1} c_{0}=1$.
(a) Under what circumstances is solution $y_{1}(t)$ a nonzero solution? Explain why $y_{2}(t)$ is a nontrivial solution.
(b) Form the function $y_{s}(t)=y_{1}(t)+s y_{2}(t)$. Here, $s$ is a constant known as the shooting parameter. Show, for any value of the constant $s$, that
$$
\begin{aligned}
&y_{s}^{\prime \prime}+p(t) y_{s}^{\prime}+q(t) y_{s}=g(t), \quad a<t<b \\
&a_{0} y_{s}(a)+a_{1} y_{s}^{\prime}(a)=\alpha .
\end{aligned}
$$