Suppose that f:R->R is twice differentiable and that f''(x) + f(x) = 0, f(0)=0 and f'(0)=0
Prove that f'(x) = f(x) = 0 for all x
The Attempt at a Solution
I can solve this using methods from calulus, using an auxillary equation and the boundary conditions. However, I am unsure how to go about it as a piece of pure maths. In examples in my notes I have needed to use the 'Identity Theorem', a corollary of the Mean Value Theorem, stating if f: (a,b) -> R is differentiable and satisfies f'(t) = 0 for all t in (a,b) then f is constant on (a,b). However, I am unsure whether this is the correct method in this case, and if it is, how to use it.