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Very basic Q about solns to y" = y

  1. Oct 6, 2015 #1
    Wolfram and the Linear Algebra text I'm currently working on, give the two possible solutions of [itex]\frac{d^2y}{dx^2}=y[/itex] as being [itex]e^{x}[/itex] and [itex]e^{-x}[/itex], or rather, constant multiples of them.

    Here wolfram agrees:


    My question is, why isn't [itex]y = e^{x} + x[/itex] and [itex]y = e^{-x} + x[/itex] not solutions?

    Or is an added polynomial in the solution just part of some nullspace of special solutions that we consider separately in general?

    edit: wouldn't the added polynomial change the dimension of the solution space?
  2. jcsd
  3. Oct 6, 2015 #2
    Why should y=e^x + x be a solution? y=x is not a solution.
  4. Oct 6, 2015 #3
    If you take y=e^x + x and diff it twice...damn it...ok, I see it now...this was dumb.
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