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TT'(t) - cT(t) = 0

  1. Oct 15, 2006 #1
    What sort of ODE is

    [tex]tT'(t)-cT(t)=0[/tex] for a real positive c?
     
  2. jcsd
  3. Oct 15, 2006 #2

    quasar987

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    linear homogeneous of first order?
     
  4. Oct 16, 2006 #3
    I tried solving it using the substitution T=tv like in the book for homogeneous equations, but it doesn't work.
     
  5. Oct 16, 2006 #4

    quasar987

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    I don't know about this trick. But when you have an ode of the form

    [tex]y'+P(t)y=Q(t)[/tex], the solution can be found by multiplying the equation by an integrating factor

    [tex]\mu=e^{\int P(t) dt}[/tex]
     
    Last edited: Oct 16, 2006
  6. Oct 16, 2006 #5
    Got it. Thanks a lot.
     
  7. Oct 17, 2006 #6
    In the original posters case, the equation is even simpler to solve since the ODE is separable.
     
  8. Oct 17, 2006 #7

    HallsofIvy

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    Also that is an "Euler-type" equation or "equipotential" equation since the coefficient of each derivative (here only one) is t to a power equal to the order of the derivative and so T= tr, for some r, is a solution.

    But, as d leet said, it is separable and easily integrable.
     
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