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Differential Equations: Finding general solution problem

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of (D^4 - I)^2(D^2 - 4D + 13I)^2(y) = 0

    2. The attempt at a solution

    My issue with this problem is that I have no clue as to what the I's mean. I have become familiar with D being used notationally with differential equations, but the introduction of the I's is totally foreign to me, and my professor has never even addressed them. Am I supposed to assume the I's are simply a constant, or am I totally missing something here?

    I can break the problem down to the following, though:
    [(D^2 + sqrt(I))(D + sqrt(I))(D - sqrt(I))]^2 * [D^2 - 4D +13I]^2 * y = 0

    From here, am I supposed to proceed "as usual" with solving the equation... or do the I's have some significance? It seems very possible to solve w/ I's being a constant, but absolutely brutal to actually find the gen. solution for :/

    Thank you for any help!
     
  2. jcsd
  3. Dec 14, 2008 #2

    CompuChip

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    I is usually used for the identity operator: I f = f.
    So for example, solving
    (D^2 - 3 I) = 0
    would mean: find all functions such that
    [tex]\frac{d^2}{dx^2} f(x) - 3 f(x) = 0[/tex].

    Also used is "id", a 1 with a hat, or a double struck 1.
    Physicists tend to leave out the I altogether, and write
    (D^2 - 3) = 0
    for the above equation.
     
  4. Dec 14, 2008 #3
    OK, thank you. So... to clarify, I can essentially ignore the I and just work with the I's coefficient as a constant, correct?
     
  5. Dec 15, 2008 #4

    CompuChip

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    Essentially, yes.
    Note however, that I is an operator, just like D is, so you cannot really write sqrt(I) any more than you can sqrt(D). You can use that I^n = I though, for any (integer) n.
     
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