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Basic differential equation

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that f(x) = A exp(σx) + B exp(-σx) is a solution to the
    following differential equation:

    f''(x) = (σ^2)f(x)

    where A, B, and σ are constants. What if a boundary condition is
    included that f(-∞) = 0?


    2. Relevant equations
    differential equation: f''(x) = (σ^2)f(x)
    solution: f(x) = A exp(σx) + B exp(-σx)

    3. The attempt at a solution
    Proof -
    Plugging exp(σx) and exp(-σx) into the equation f''(x) = (σ^2)f(x) gives an equality, therefore any linear combination of exp(σx) and exp(-σx) is a solution.


    If the boundary condition f(-∞) = 0 is introduced, would only exp(σx) be a solution since exp(-∞)=0?
     
  2. jcsd
  3. Jan 10, 2013 #2

    haruspex

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    That certainly allows B=0 to be a solution, but show also that B must be zero.
    (Of course, this is assuming sigma > 0.)
     
  4. Jan 10, 2013 #3
    I'm confused about how to do this?

    B must be zero because Bexp^(-∞) = 0... is that sufficient?
     
  5. Jan 11, 2013 #4

    SammyS

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    No. e-σx → +∞ as x → -∞ , if σ > 0 .
     
  6. Jan 12, 2013 #5
    So if the boundary condition f(-inf) is involved, then is the answer Aexp()???

    Can someone please explain boundary conditions?
     
  7. Jan 12, 2013 #6

    tiny-tim

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    hi cytochrome! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes, the general solution is y = Aeσx
    to completely solve a differential equation, you need as many boundary conditions as there are constants

    here, there were two constants but only one boundary condition, so your solution still has one unknown constant
     
  8. Jan 12, 2013 #7
    Thanks!

    So the question is for the boundary condition f(-∞), therefore the answer would just be the general solution?
     
  9. Jan 12, 2013 #8

    haruspex

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    Your very first solution was correct, it's just that the reasoning you offered was insufficient. A exp(σx) is the solution that satisfies the boundary condition. You just have to prove that B=0. What would happen to f(x) at -∞ if B were nonzero?
     
  10. Jan 13, 2013 #9
    If B ≠ 0 then f(-∞) = 0 doesn't make sense since this only works if f(x) = Aexp(σx).

    Is my reasoning correct?
     
  11. Jan 13, 2013 #10

    HallsofIvy

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    I don't see any "reasoning" here, simply the assertion that "if B ≠ 0 then f(-∞) = 0 doesn't make sense". WHY does it not make any sense?
     
  12. Jan 13, 2013 #11
    Well f(-∞) = 0 implies that Aexp(x) is the only possible answer. Bexp(-x) doesn't work because as x → -∞ then Bexp(-x) → 0

    Is that a good way to show it?
     
  13. Jan 13, 2013 #12

    haruspex

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    No it doesn't! Get that right and you're home.
     
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