# Analysis - solutions to differential equations

1. Apr 18, 2010

### Kate2010

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

Thanks :)

2. Apr 18, 2010

### snipez90

Consider the derivative of the function $g = (f')^2 + f^2,$ apply the "Identity Theorem", and use the initial conditions.

3. Apr 19, 2010

### Kate2010

Thanks! That makes a lot of sense.

The next part of the question is a more general version: If g is twice differentiable and satisfies h''(x) + h(x) = 0 prove that h(x) = Acosx +Bsinx