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Solving a First Order Differential Equation with Initial Conditions.

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem:

    t(dy/dt)+8y=t^3 where t>0 and y(1)=0

    2. Relevant equations

    None?

    3. The attempt at a solution

    It's a linear equation, so rearranged to dy/dt+8y/t=t^2.

    Took the integrating factor e^(∫8/tdt)=t^8 and multiplied through.

    (t^8)dy/dt+8t^7y=t^10.

    ∫(t^8y)'dt=∫t^10dt
    t^8y=(t^11)/11 + C

    Solve for C, I got -1/11.

    Final solution is y=(t^3)/11 - 1/(11t^8)

    This isn't right, though. Does anyone see where I made the mistake?

    Thanks!
     
  2. jcsd
  3. Jan 17, 2013 #2

    Mark44

    Staff: Mentor

    Your answer is correct. Are you comparing with the book's answer? As long as you are working the same problem, your answer checks.
     
  4. Jan 17, 2013 #3
    Hmm, strange. Maybe a typo. Thanks!
     
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