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Differential Equations and Initial Values

  1. Feb 14, 2013 #1
    1. The problem statement, all variables and given/known data
    If y is the solution of the initial value problem

    y'(t) + y(t)/t = 1/t4

    y(2)=1

    What is y(3)

    2. Relevant equations



    3. The attempt at a solution

    I u(t) = e∫1/t dt = eln t = t

    ∫[yt]' dt = t-4 dt

    yt = -1/3t3 + C

    Plugging in my initial value I solve for C:

    1 = -1/3(24) + C/t

    C = 2 + 1/3(23)

    C = 2 + 1/24 = 49/24

    y(3) = -1/3(34) + 49/24

    y(3) = 3961/1944

    The answer should be 149/216 but I can't see where I'm going wrong. Help is appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 14, 2013 #2

    mfb

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    I have no idea what you are doing here.

    Let's check:
    I think "yt" should be y(t)? And the t3 is part of the denominator?
    Like this: y(t) = -1/(3t3) + C
    y'=1/t4
    Therefore, ##\frac{1}{t^4} -\frac{1}{3t^4} + \frac{C}{t} = \frac{1}{t^4}##
    This cannot be true for all t.
     
  4. Feb 14, 2013 #3

    LCKurtz

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    That step is incorrect. First write the correct equation ##(ty)' = ?##, then integrate it.
     
  5. Feb 15, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    He's finding the "integrating factor" for a linear differential equation. But, as LCKurtz says, he used it improperly. Multiplying both sides of y'(t) + y(t)/t = 1/t4 by t gives
    ty'(t)+ y(t)= (yt)'= 1/t3
    No, it clearly should not be: integrating (ty)' gives t times y.

     
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