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Analysis - solutions to differential equations

  1. Apr 18, 2010 #1
    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. jcsd
  3. Apr 18, 2010 #2
    Consider the derivative of the function [itex]g = (f')^2 + f^2,[/itex] apply the "Identity Theorem", and use the initial conditions.
     
  4. Apr 19, 2010 #3
    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
    Using your advice, I can show h'(x)2+h(x)2 = constant
    I see that this looks a bit like Pythagoras but am not sure how I would prove that it involves sin and cos.
     
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