Linear shooting hits the far boundary in one sensitivity step

Consider the linear two-point problem \[ y''=y-t,\qquad y(0)=0,\qquad y(1)=0. \] 1. Solve the initial-value problem with $y(0)=0$ and $y'(0)=s$ in closed form. Deduce the unique missing slope $s_*$ for which $y(1;s_*)=0$, and write the exact solution $y(t)$. Equivalently, introduce the pair of IVPs \[ \begin{aligned} u''&=u-t,& u(0)&=0,& u'(0)&=0,\\ v''&=v,& v(0)&=0,& v'(0)&=1, \end{aligned} \] and obtain $s_*$ from $u(1)+s_* v(1)=0$. 2. Let $z=\partial y/\partial s$. Derive the sensitivity IVP satisfied by $z$, evaluate $z(1)$, and perform one Newton update \[ s_1=s_0-\frac{y(1;s_0)-0}{z(1;s_0)} \] from the initial guess $s_0=0$. Prove that $s_1=s_*$ and explain why a linear BVP needs only one such correction. 3. Now discretize the first-order system $y'=v$, $v'=y-t$ by Heun's method (improved Euler) with $h=1/2$, starting from $y_0=0$, $v_0=s$. Compute the discrete map $s\mapsto y_2(s)$ with exact rational arithmetic, and solve $y_2(s)=0$. Confirm that one discrete Newton step from $s_0=0$, using the discrete sensitivity $\partial y_2/\partial s$, recovers the same slope. 4. Do not replace shooting by a centered three-point Poisson stencil, do not treat the far condition as $y'(1)=0$, and do not iterate Newton as if $s\mapsto y(1;s)$ were a genuinely nonlinear scalar map.

A particular solution of $y''=y-t$ is $y=t$. The homogeneous solutions are $\sinh$ and $\cosh$, and $y(0)=0$ kills the $\cosh$ coefficient.