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Eigenvectors of first-order differential equation

  1. May 17, 2009 #1
    Suppose that f is an eigenvector of T with corresponding eigenvalue [tex]\lambda[/tex]. Then f' = T(f) = [tex]\lambda[/tex]f. This is a first-order differential equation whose solutions are of the form f(t) = ce^([tex]\lambda[/tex]*t) for some scalar constant c. Consequently, every real number [tex]\lambda[/tex] is an eigenvalue of T, and [tex]\lambda[/tex] corresponds to eigenvectors of the form ce^([tex]\lambda[/tex]*t ) for c not equal to 0. Note that for [tex]\lambda[/tex] = 0, the eigenvectors are the nonzero constant functions.

    Question: I was wondering if someone could explain the following sentence from above: "This is a first-order differential equation whose solutions are of the form f(t) = ce^([tex]\lambda[/tex] * t) for some scalar constant c. "

    Work:
    f ' = T(F) = [tex]\lambda[/tex]f.
    Therefore, f ' = [tex]\lambda[/tex] f ==> [tex]f' / f [/tex] = [tex]\lambda[/tex]
    So, [Integral] [tex]f ' / f [/tex] dt = [Integral] [tex]\lambda[/tex] dt
    Therefore, ln(f) = [tex]\lambda[/tex]t + c
    <==> e^( ln(f) ) = e^( [tex]\lambda[/tex]t + c )
    = [tex]e^l^n^|^f^|[/tex] = e ^ ( [tex]\lambda[/tex]t + c )
    = [tex]e^c[/tex]e^( [tex]\lambda[/tex] * t )
    <==> f = [tex]e^c[/tex]e^( [tex]\lambda[/tex] * t ) which does not equal ce^( lambda * t ) ??

    Question: I'm kind of rusty with calculus but I didn't think [Integral] [tex]f ' / f [/tex] dt = ln(f)
    I thought it would be [Integral] [tex]f ' / f [/tex] dt = [Integral] [tex] 1 / f [/tex] dt = [tex]t / f[/tex]
     
    Last edited: May 17, 2009
  2. jcsd
  3. May 18, 2009 #2

    tiny-tim

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    Hi jeff1evesque! :smile:

    (have a lambda: λ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    Yes it does … the constant ec has simply been renamed c :wink:
    ∫f/'f dt = ln(f) + c because d/dt (ln(t)) = 1/t, so by the chain rule d/dt (ln(f(t))) = f'(t)/f(t) :wink:
     
  4. May 18, 2009 #3
    Wow, thanks a lot Tim.

     
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