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Little bit confuse on vector space

  1. Apr 23, 2009 #1
    how to proof if the solution set of a second order diffential equation af''+bf'+cf=0 is a real vector space w.r.t. the usual opeations?
     
  2. jcsd
  3. Apr 23, 2009 #2

    jambaugh

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    Since the set of differentiable functions is itself a vector space the solutions would form a subspace. It thus is sufficient to show that the set is closed under the operations of addition and scalar multiplication.

    Given any subset of a vector space you already have all the properties of associativity, distribution under scalar multiplication and vector addition, etc. The only issue is closure under the basic operations.
     
  4. Apr 23, 2009 #3
    james,

    first of all thank you very much for your explaining, that helps me a lot.

    but still i have some question to ask you, can you please help me out as well?

    i just got no idea what is the set of the soluiton of those d.e.
    do i need to use y=a^ex to solve them or i need to reduce them into first order matrix system?

    but if i reduce into first order matrix system, how can i proof it is closeure under addition and scalar multiplication?

    many thanks!!

    regards,
    tony
     
  5. Apr 23, 2009 #4

    jambaugh

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    Either method works for finding solutions but solving the system directly is the most straightforward (presuming a,b, and c are constants). The way to look at this equation is in terms of the derivative as an operator D:

    [aD^2 +bD + c1] f = 0
    The exponential function f=e^(rx) is an "eigen-vector" of the D operator with eigen-value r. The set of all such function forms an "eigen-basis" so any solution must be a linear combination of exponential functions and you can find the r's algebraically.
     
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