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Differential Equations: Find the Solutions and Limiting Factos

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the solution and limiting factor

    (t+1)y' + y = 6

    y(1) = -2


    2. Relevant equations



    3. The attempt at a solution

    So I started off by isolating y'

    y' + y/(t+1) = 6(t+1)

    Then I found u(t)

    u(t) = e∫ t+1 dt = et2/2 +t

    y' =(et2/2 +t) = y(et2/2 +t) = 6(et2/2 +t)/(t+1)

    The difficulty I'm having is integrating 6(et2/2 +t)/(t+1)

    u-substitution doesn't help. Suggestions are appreciated
     
  2. jcsd
  3. Feb 4, 2013 #2

    LCKurtz

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    That should be ##u(t) = e^{\int \frac 1 {t+1}}\, dt##.
     
  4. Feb 4, 2013 #3
    Ahh, thank you.

    Can you explain to me what is meant by the limiting factor? Is it the largest range of t values for the given information of y(1) = -2?
     
  5. Feb 4, 2013 #4

    LCKurtz

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    I am not familiar with the term "limiting factor". Perhaps it refers to disallowed values of t in your answer.
     
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