1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:

    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?

  2. jcsd
  3. Oct 23, 2012 #2


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

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


    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


    User Avatar
    Homework Helper

    You would want the solution corresponding with the positive initial condition, so yes you ignore the negative case.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook