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Linear equation (differential equations) problem

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve:

    dP/dt + 2tP = P + 4t - 2


    3. The attempt at a solution

    I've done a couple of these but I'm not sure how to start with this one...

    First I have to put it in standard form right?

    dP/dt + P(2t - 1) = 4t - 2

    Then obtain the integrating factor by integrating (2t - 1)
     
  2. jcsd
  3. Jan 20, 2013 #2

    LCKurtz

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    Right so far. Now calculate$$
    e^{\int 2t-1 dt}$$for your integrating factor.
     
  4. Jan 20, 2013 #3
    Ok! I get:

    e^(t^2 - t)

    Then we include it in the equation and integrate both sides?

    d [e^(t^2 - t) P ] / dt = e^(t^2 - t) * (4t - 2)

    The right hand side integral is giving me problems though...
     
  5. Jan 20, 2013 #4

    LCKurtz

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    Try a u substitution: ##u = t^2-t## as your first step.
     
  6. Jan 20, 2013 #5
    Oh god, I can't believe I missed that one hehe I was overcomplicating it...

    I end up with:

    P = [2 e^(t^2 - t) + C ] / e^(t^2 - t)

    Interval of definition would be (-inf, inf) correct?
     
  7. Jan 20, 2013 #6

    LCKurtz

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    Yes, since the denominator is never zero.
     
  8. Jan 20, 2013 #7
    Thanks a lot!
     
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