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Wronskian to prove linear independence

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]v_1,v_2[/itex] be any two solutions of the differential equation [itex]y''+ay'+by=0[/itex] such that [itex]\frac {v_2}{v_1}[/itex] is not constant, and let f(x) be any solution of the differential equation as well.

    Use the properties of the Wronskian to prove that constants [itex]c_1,c_2[/itex] exist such that:

    [tex]c_1 v_1(0) + c_2 v_2(0) = f(0), \qquad c_1 v_1 '(0) + c_1 v_1 '(0) = f' (0)[/tex]


    2. Relevant equations
    Here are the relevant properties of the Wronskian, defined as [itex]W(x)=v_1(x) v_2 '(x) - v_2(x)v_1 '(x)[/itex]:

    Let W be the Wronskian of two solutions [itex]v_1, v_2[/itex] of the differential equation [itex]y'' + ay' +by =0[/itex].
    All the following holds:
    [tex]W' +aW =0[/tex]
    [tex]W(x) = W(0)e^{-ax}[/tex]
    [tex]W(0) = 0 \iff \frac{v_2}{v_1} \text{is constant}[/tex]

    3. The attempt at a solution

    [itex]\frac{v_2}{v_1}[/itex] is not constant, so [itex]W(0) \ne 0[/itex], and therefore for some constant [itex]d[/itex] we have
    [tex]dW(0)=f(0)[/tex]
    [tex]d(v_1(0) v_2 '(0) - v_2(0)v_1 '(0)) = f(0)[/tex]
    [tex][dv_2'(0)]v_1(0) + [-dv_1'(0)]v_2(0) = f(0)[/tex]

    So for our solution, [itex]c_1 = dv_2'(0)[/itex] and [itex]c_2 = -dv_1'(0)[/itex], but this leads to

    [tex][dv_2'(0)]v_1'(0) + [-dv_1'(0)]v_2'(0) = f'(0)=0[/tex]

    Which is not always true.
     
  2. jcsd
  3. Sep 22, 2011 #2
    Doh! - nevermind. Is there a way to delete a post?
     
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