Recipe

Solve a linear BVP by shooting

Use when: you have a LINEAR second-order BVP $y''=p(x)y'+q(x)y+r(x)$ , $y(a)=\alpha$ , $y(b)=\beta$ , with $p,q,r$ continuous on $[a,b]$

Goal: the BVP's solution values at every grid point, accurate to whatever error RK4 itself carries -- reached by two ordinary IVP solves and one exact combination, no iteration

Taught in The Linear Shooting Method

Quick recipe

Reduce the equation to a first-order system
Shoot once with the full equation, guessed slope $0$
Shoot again with the homogeneous equation, guessed slope $1$
Combine with $c=(\beta-y_1(b))/y_2(b)$

1
Write $u_1=y,\ u_2=y'$ , exactly Chapter 12 Section 11's own move: $u_1'=u_2,\ u_2'=p(x)u_2+q(x)u_1+r(x)$ .

Example

For $y''=y+x$ (so $p=0,\ q=1,\ r(x)=x$ ) on $[0,1]$ : $u_1'=u_2,\ u_2'=u_1+x$ .

Why: RK4, and every method Chapter 12 built, only ever knows how to advance a full state $(u_1,u_2)$ one step. This is what makes "integrate forward" possible at all.
2
Solve $u_1(a)=\alpha,\ u_2(a)=0$ forward to $b$ with RK4, keeping the FULL $r(x)$ . Call the resulting solution values $y_1(x)$ .

Example

$\alpha=1$ : RK4 with $h=0.25$ gives $y_1(0)=1,\ y_1(0.25)\approx1.03402,\ y_1(0.5)\approx1.14870,\ y_1(0.75)\approx1.36696,\ y_1(1)\approx1.71821$ (the exact solution of this particular IVP is $y_1(x)=e^x-x$ , so this already matches $e^1-1\approx1.71828$ to 4 places).

Why: this shot already satisfies the LEFT boundary condition exactly ( $u_1(a)=\alpha$ by construction), and solves the real equation. It just has no reason yet to land on $\beta$ at $b$ , since the slope $0$ was arbitrary.
3
Solve $u_1(a)=0,\ u_2(a)=1$ forward to $b$ with RK4, this time with $r(x)$ replaced by $0$ . Call the resulting values $y_2(x)$ .

Example

Homogeneous: $u_2'=u_1$ (no $+x$ term). RK4 with $h=0.25$ gives $y_2(0)=0,\ y_2(0.25)\approx0.25260,\ y_2(0.5)\approx0.52108,\ y_2(0.75)\approx0.82229,\ y_2(1)\approx1.17516$ (exactly $\sinh x$ here, so $y_2(1)\approx\sinh1\approx1.17520$ ).

Why: zeroing $r$ is what makes $y_2$ purely a "sensitivity" measurement. $cy_2$ can be added to $y_1$ without introducing a second, unwanted copy of $r(x)$ into the combined solution.
4
Compute $c$ from the two numbers Steps 2-3 just produced, then report $y(x)=y_1(x)+cy_2(x)$ at every grid point.

Example

$c=\dfrac{1-1.71821}{1.17516}\approx-0.61116$ . Combined: $y(0.25)\approx1.03402+(-0.61116)(0.25260)\approx0.87964$ , and by construction $y(1)=1.71821+(-0.61116)(1.17516)\approx1.00000$ -- exactly $\beta$ .
- $y_2(b)$ is (numerically) zero — no unique blend exists. The matching homogeneous BVP has a solution that isn't just zero -- exactly what Section 1's theorem's $q(x)>0$ hypothesis rules out. In practice this rarely shows up as a clean division error. $y_2(b)$ is usually only CLOSE to zero, not exactly zero (RK4's own round-off sitting where the true value is exactly $0$ ). The visible symptom is $c$ swelling to a huge number that swings wildly with the step count, rather than a clean failure.
- $y_2(b)\ne0$ — $c$ is well-defined and unique; proceed as above.
Why: superposition (the Textbook tab's own derivation) guarantees $y_1+cy_2$ solves the equation and the left condition for every $c$ . This step's algebra is the ONLY place the right condition ever gets imposed, and it does so exactly, not approximately.

Common mistakes

Trying this exact combine-and-done trick on a NONLINEAR equation. Superposition is what makes Step 4's single division exact, and superposition is a fact about LINEAR equations only. A nonlinear $r$ (or nonlinear $p,q$ depending on $y$ ) means $y_1+cy_2$ generally doesn't even solve the original equation for $c\ne0,1$ . There is no algebraic $c$ to solve for. Section 3's genuinely different method (Newton's method on the boundary miss, reusing Chapter 11's own recipe by name) is what a nonlinear BVP actually needs.
Forgetting to zero out $r(x)$ in Step 3. Running the SAME nonhomogeneous equation for both shots breaks superposition: $y_1+cy_2$ would then solve $y''=py'+qy+(1+c)r$ , not the original equation, for every $c$ except $c=0$ .
Swapping which shot gets which starting values. Step 2's slope is $0$ (not $\alpha$ twice), and Step 3's height is $0$ (not $\alpha$ again). Using $\alpha$ for both shots' starting heights makes $y_2$ satisfy the SAME left condition as $y_1$ , instead of measuring pure sensitivity to slope -- and the combination's left boundary condition breaks.
A sign slip in Step 4. Computing $c=(y_1(b)-\beta)/y_2(b)$ instead of $(\beta-y_1(b))/y_2(b)$ flips the correction's direction. It overshoots the target by twice the original miss, instead of closing the gap.