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Finding an interval in which the IVP has a unique solution.

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is the question with the solution from the textbook:
    8IaP6.png

    I don't get how looking at (2/t) and 4t tells us that the solution must be on the interval 0 to infinity. Don't we have to set y' to zero and solve for how the directional fields behave and thus find out how y behaves as t -> infinity?

    I don't think I am visualizing this correctly and thus I don't understand how the answer works. Could anyone please explain?

    Thanks
     
  2. jcsd
  3. Oct 23, 2012 #2

    Zondrina

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    Notice your initial condition, y(1) = 2, which is positive, yes? This corresponds to your interval of t being strictly positive ( Since if t = 0, then our equation is undefined ).

    So 0 < t < ∞
     
  4. Oct 23, 2012 #3
    So because of the discontinuity at t = 0, the solution will never be negative if the initial condition is positive?
     
  5. Oct 23, 2012 #4

    Zondrina

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    Yes indeed.
     
  6. Oct 23, 2012 #5
    What about this question? Why is the negative answer from taking the root of y^2 rejected?

    X4FUF.png

    Is it because y has a discontinuity at 0 and thus because our initial condition y(2) = 3 starts in the positive y the solution must vary between 0 < y < infinity?

    I just want to clarify this one because it is slightly different when compared to my original question in that the discontinuity is y instead of t or x.

    Thank you!
     
  7. Oct 23, 2012 #6

    Zondrina

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    You would want the solution corresponding with the positive initial condition, so yes you ignore the negative case.
     
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