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Homework Help: Finding The Solution of A Differential Equation

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    y' + 3y = t + exp(2t)

    2. Relevant equations

    3. The attempt at a solution
    My solution is y = t/3 + exp(-2t) -1/9 + c. The solution in the back of the book is y = cexp(-3t) + (t/3) - (1/9) + exp(-2t). How come the two solutions differ so vastly? I followed the procedure outlined in the textbook precisely.
    Last edited: Sep 2, 2013
  2. jcsd
  3. Sep 2, 2013 #2


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    The solution I'm getting is :

    ##y(t) = \frac{c}{e^{3 t}} + \frac{t}{3} + \frac{e^{2 t}}{5} - \frac{1}{9}##

    You should check your work over again I think. You may have made a slight error.
    Last edited: Sep 2, 2013
  4. Sep 2, 2013 #3
    I got the exact same answer. I attached my work as a file. According to my text book, the solution to this sort of differential equation is:

    [itex]y = \frac{1}{\mu (t)} \int \mu (t) g(t) dt + c[/itex], where [itex]\mu (t) = e^{\int p(t)dt}[/itex]

    Attached Files:

    • DE.jpg
      File size:
      14 KB
  5. Sep 2, 2013 #4


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    Yes your book is 100% correct ( Barring I'd leave constants all the way until the end ), but don't neglect the ##e^{3t}## in the denominator. It will vary the value of ##c## if you're ever given an initial value problem.
    Last edited: Sep 2, 2013
  6. Sep 2, 2013 #5
    Oh, I see. So exp(-3t) would multiply the c, after I did the integration.
  7. Sep 2, 2013 #6


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    It will divide it actually. After evaluating the integrals I got this :

    ##e^{3t}y(t) = e^{3t} ( \frac{e^{2t}}{5} + \frac{t}{3} - \frac{1}{9} ) + c##
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