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Which is the correct expression for the Differential Equation

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Which of the following is a correct expression for

    1t(s-1)y'(s) ds

    I know the answer is:

    a) 1/2(t-1)2y(t)

    b) 1/2(t-1)2y'(t)+ (t-1)(y(t)-y(1))

    c) (t-1)(y(t)-y(1))

    d) (t-1)y(t)-∫y(s) ds



    2. Relevant equations



    3. The attempt at a solution

    I know the answer is d, but I don't understand why. Are they integrating the given equation, and if so, how do you integrate a function without knowing what the function is?

    I have attached an image of the original question
     

    Attached Files:

  2. jcsd
  3. Feb 10, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    You can do integration by parts - the remaining integral is just an integral over y(s), and you can find that in answer (d). The other part can be evaluated without an integral.
     
  4. Feb 10, 2013 #3

    Dick

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    Science Advisor
    Homework Helper

    You'll notice they didn't actually integrate it in choice d). The just expressed the original integral in terms of another integral. The technique is called 'integration by parts'. It should be described in your course.
     
  5. Feb 10, 2013 #4
    I think I've got it now. It's been a while since I've done integration by parts. Thanks for the input.
     
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