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Finding the general solution of a first order diff eq with a variable coefficient

  1. Oct 2, 2011 #1
    I have the following two questions to solve

    Problem 1.

    3x' + 1/t x = t

    and

    Problem 2.

    x' + 1/t x = ln t

    I have followed a method detailed in my textbook to try and get an answer for Problem 1 but am a bit unsure so if anyone can clarify my workings below before I spend time trying to solve Problem 2.

    3x' + 1/t x = t

    Fits the format dx/dt + g(t)x = f(t)

    For the integrating factor I(t) = e^∫g(t) dt

    ∫1/t = ln t

    e^ln t = t

    Multiply both sides by I(t) so the equation becomes d/dt(I(t)x(t)) = I(t)f(t)

    3 d/dt (tx) = t^2

    Then I(t)x(t) = ∫I(t)f(t) dt + C

    3tx = ∫t^2 dt + C

    3tx = (t^3)/3 + C

    x(t) = ((t^3)/3 + C)/3t

    x(t) = 1/6t^2 + C1/3t^-1

    Does this look right? If so I will attempt Problem 2. Thanks for any assistance.
     
  2. jcsd
  3. Oct 2, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    3*3 is NOT 6! Other than that, and the need for parentheses to make it clearer (most people would interpret "1/6t^2" as "1/(6t^2)" and I don't think that is what you mean), you are doing this correctly.

     
  4. Oct 2, 2011 #3
    You are right my parentheses does need to be clearer.

    My final answer should be:

    x(t) = 1/9(t^2) + C1/3(t^-1)

    Does that look better?
     
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