I have the BVP : $y'' + k^2y = f(x), y(0) = 2, y(1) = 5$, and I need to find values for k such that it has a unique solution.
I know that this happens if the associated homogeneous equation $y'' + k^2y = 0$ has only the trivial solution. The solution to this is $y_h = Acos(kx) + Bsin(kx)$. So I need to find values of k such that $y_h = 0$ for all of them? Applying the boundary conditions returned $A = 2$ and $B = (5-2cos(k))/sin(k)$, but using this expression for B only seems to complicate things. Can anybody give me a hint? Thanks!
All you need is $\sin k = 0$. As you have shown there exist values for $A$ and $B$ which give a non-zero solution when $\sin k \neq 0$. So the condition on $k$ is $k =n\pi$ for some integer $n$. Note that when $\sin k=0$ we get $y_h=A\cos kx$ and \the second condition is not satisfied in this case (since we cannot have $2\cos k =5$).