Uniqueness of the Solution of The Second order Differential Equation

Question

I know that $a\cos(ct)＋b\sin(ct)$ is the general solution of $f''(t)+c^{2}f(t)＝0$, where $f$ is a twice continuously differentiable function on $\mathbb{R}$, but how can we prove the uniqueness of it? （PS:Don't use The general theorem, e.g. the existence and uniqueness theorem.）

Answer 1

Let $f_1$ and $f_2$ be 2 linearly independant solutions of the equation: $f''(t) = -c^2.f(t)$. We can also write the equation in terms of the laplacian operator as: $\nabla ^2 {f} = -c^2.f$. Now consider a region bounded by some curve $C$. The value of $f$ at some point $P \in C$ is $a$. Since $f_1, f_2$ are both solutions, they must satisfy the differential equation:

$\nabla ^2{f_1} = -c^2.f_1$..........1)

$\nabla ^2{f_2} = -c^2.f_2$..........2)

Also, $f_1(P) = f_2(P) = a$...........3)

From 1 and 2, by subtraction we get:

$\nabla ^2{f_1 - f_2} = -c^2.(f_1-f_2)$

Thus, $g = f_1 - f_2$ must also be a solution. This means $g(P) = a$ as well. But clearly, $g(P) = 0$. Contradiction. Hence there exists a unique solutiom satisfying the differential equation and some boundary value.