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1st order linear diff equation clarification

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    determine the largest interval on which a unique solution exists:

    y' + (t/((t^2)+1)) y = sin t with initial conditions: y(-2) = 1

    2. Relevant equations
    arent there discontinuities at t = 0 and t = -1?
    (textbook says no, based on answer..i am confused why not?) i think it may have something to do with not understanding how the initial conditions, eg, y(-2) = 1 relate to the interval.

    3. The attempt at a solution
    the equation is already in the form y' + p(t)y = g(t)
    p(t) = (( t/(t^2 ))+1))

    if there are discont @ t = 0 and t= -1, then the intervals would be:
    (- infin, -1)
    (-1, 0)
    (0, infin)

    therefore largest interval for t = -2 is: - infin < t < -1

    the answer in back though says largest interval is: - infin < t < infin

    please help, thanks alot.

    EDIT: i glossed over the square of t, so i now know that t= -1 does not make p(t) = 0.. but i am still unclear as to why t = 0 is not a discontinuity.. Also, i think i am unclear about what precisely is a discontinuity. thanks again for any help..
    Last edited by a moderator: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2


    Staff: Mentor

    Your function p(t) is defined and continuous for all real numbers. Why would you think there is a problem at t = 0?
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