Intro
Simple shooting methods for boundary value problem of the form: $$ x'(t)=f(x(t),t), \quad r(x(a),x(b))=0 $$ with $f:\Omega \times I \rightarrow\mathbb{R}^m, \quad r: \mathbb{R}^m \times \mathbb{R}^m \rightarrow \mathbb{R}^m, \quad I=[a,b]$
try to find a starting vector $s$ so that the solution for $$ x'(t)=f(x(t),t), \quad x(a)=s $$ provides a solution for $x(t)$ with $r(x(a),x(b))=0 $. The first step for that is that if $x(t;s)$ is the solution for $x(t)$ of $x'(t)=f(x(t),t), \quad x(a)=s$ with start value $x(a)=s$, we have to find the roots of the function $$ F(s):= r(s;x(b;s)) $$
Task
I have given the system $$ x''(t)=\gamma x, \quad x(0)=1, \quad x(1)=2 $$ and need to formulate $F(s)$.
My try
I am absolutely clueless. The only thing I would say ist that $a=0, b=1$ but from there on I don't know what to do. Do I really need to simply put down $$ F(s):= r(s;x(b;s))= r(s;x(1;s)) $$ I cannot imagine that it would be this simple. I don't know how to go further, only using Wikipedia I would assume that $$ F(s)=x(1;s)-2 $$ would be the solution but I don't really know if this is right or why. I would appreciate your help!
In the full formalism as given you have $s=(s_1,s_2)=(x(0),x'(0))$ and $F(s)=(x(0;s)-1,\,x(1;s)-2)$.
Of course you can reduce that boundary function to a scalar function by always setting $s_1=1$. But this is a less general formulation.