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Solving a First Order Initial Value Problem

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    If y = y(t) is the solution of the initial value problem
    y' + (2 t + 1) y = 2 cos(t)
    y(0) = 2
    What is y''(0)?

    2. Relevant equations


    3. The attempt at a solution
    Since this is a first order linear, I started out by finding the integrating factor so I can find what y is, and then just take the second derivative of it, and then put it to 0.

    [itex] IF = e^{ \int {2t + 1} } => IF = e^{t^2 + t} [/itex]
    Then, I end up with this equation, if we set IF = k.
    [itex] (ky)' = k (2 cos(t)) [/itex]

    From here, I don't think I can integrate that equation, so now I'm pretty much stuck. Is there a method I'm missing? Or could I find the value of the second derivative using the value of y at 0?
    Any help is appreciated, thank you.
     
  2. jcsd
  3. Nov 2, 2014 #2

    LCKurtz

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    Instead of trying to solve the original DE, try just differentiating it once and solving for ##y''## and see if you can get the answer from that.
     
  4. Nov 2, 2014 #3
    Wow! I can't believe I didn't see that, haha, thanks man. Much appreciated.
     
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