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The solution to the Differential Equation

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The Initial value problem:

    y' - (3/t)y = 0

    y(1) = -10

    Has the solution:

    I have attached an image of the question


    2. Relevant equations



    3. The attempt at a solution

    Firstly I found the integrating factor:

    u(t) = e∫-3/t dt

    u(t) = -3/t

    I plugged this back into the original equation:

    (-3/t)y' + (9/t2y = 0

    ∫[-(3/t)y]' dt = ∫ 0 dt

    (-3/t)y = C

    y = -Ct/3

    But I don't understand where I go from here. The possible answers all include t3 and I don't understand where that has come from. What step did I miss?
     

    Attached Files:

  2. jcsd
  3. Feb 9, 2013 #2
    What is the integration of 1/t?
     
  4. Feb 9, 2013 #3
    Yes, I see my mistake now.

    u(t) = t^-3
     
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