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Homework Help: Find a solution to the initial value problem

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a solution to the initial value problem
    that is continuous on the interval 1d5856a507737b03db7c79778851561.png where

    2. Relevant equations
    I know the equations, but don't want to type them out.

    3. The attempt at a solution

    I got the first part of this question. The part where g(t) = sin(t)

    I can not figure it out when g(t)=-sin(t)
    The answer I have come up with for that part is y(t)= -1+(8*e^(cos(t))/e)
    This is using y(0)=y(2pi)

    Any reason I can't do this? There must be some reason because I'm not getting the right answer.
    Last edited by a moderator: Apr 16, 2017
  2. jcsd
  3. Jan 29, 2013 #2


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    Homework Helper

    Okay so I'll assume you've actually solved y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

    I got y(t) = cecos(t) + 1. As the general solution.

    As for solving the second one, I got y(t) = cecos(t) - 1 for π ≤ t ≤ 2π.

    I'm not sure how you're getting what you got?
  4. Jan 29, 2013 #3
    I have solved for y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

    Since we are given y(0)=7 we are not allowed to put C in our answer. SO my First answer was y=1+6/e*e^cos(t)

    but since we are not given something for the other interval I don't know how to solve for C. I thought I could use y(0) = 7 since 2pi and 0 are equal for sin and cos, but it is not right.
  5. Jan 29, 2013 #4


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    Your first answer is not correct, you may want to check it again, I'll assume it's an arithmetic error.

    Also, you have the right idea evaluating the second one at 2π, once again though it's probably just arithmetic.
  6. Jan 29, 2013 #5
    I'm not sure how my first is wrong? This is for online homework and I put it in and it says it is correct.
  7. Jan 29, 2013 #6


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    Yes, y=1+(6/e)*e^cos(t) is correct for the first part. Use that to find y(pi) and use it as an initial value for the second. And no, you can't assume y(0)=y(2pi). It's true for the individual parts. But it's not necessarily true if you are splicing them together and you cross the point where you joined them.
  8. Jan 29, 2013 #7


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    Gold Member

    I tried a quick and dirty Excel brute force solution with just 10 points in each interval.

    Wonder how off I am?

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