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Help with IVP

  1. Jun 11, 2007 #1


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    1. The problem statement, all variables and given/known data


    Find the solution of the DE that passes through the point (-1, -1).

    2. Relevant equations

    3. The attempt at a solution



    integrate both sides:

    [tex] ln(y) = sin(t) + C [/tex]

    Normally I would plug in -1 for y and t and solve for C but I cant take the LN of -1. When I try to isolate y first then plug in, I get the same problem with t. How can I solve this? im stuck! Is it that the ln(y) is actually ln(|y|) ??????
  2. jcsd
  3. Jun 11, 2007 #2


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    Homework Helper
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    You figured it out for yourself in the end:smile::

    Last edited: Jun 11, 2007
  4. Jun 11, 2007 #3


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    You could do the absolute value, since ln(-y)'=ln(y)=1/y*y'. Or you could just exponentiate your solution to get y=exp(sin(t)+C)=D*exp(sin(t)). Now you can put D negative.
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