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2nd Order ODE Contradiction ?

  1. Aug 6, 2013 #1
    2nd Order ODE "Contradiction"?

    To solve a 2nd order ODE, we can follow the steps as shown below. (Image 2 is a continuation from Image 1, apologies for the size difference.) :rolleyes:


    The method to obtain the solution is straightforward.

    Let's say


    If k = -1, a possible solution is y = sin x. If k = 1, a possible solution is y = e^x.

    How do we obtain these two different solutions from one straightforward method?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 6, 2013 #2
    I think Laplace transforms look good for this.

    $$\frac{d^2y}{dx^2}=ky\implies\mathcal{L}\{\frac{d^2y}{dx^2}\}(s)=k \mathcal{L}\{y\}(s)\\ s^2\mathcal{L}\{y\}(s)-\left.\frac{dy}{dx}\right|_0-sy(0)=k\mathcal{L}\{y\}(s) \\ \mathcal{L}\{y\}(s)=\frac{\frac{dy}{dx}|_0+sy(0)}{s^2-k}\\y(x)=\mathcal{L}^{-1}\left\{\frac{\frac{dy}{dx}|_0+sy(0)}{s^2-k}\right\}(x).$$
    With a bit of simplifying, this comes down to ##y(x)=c_1e^{\sqrt{k}x}+c_2e^{-\sqrt{k}x}##.

    Note that ##\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}##. :wink:
  4. Aug 6, 2013 #3
    That's pretty neat.
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