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Differential Equations

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the exact solution of the initial value problem. Indicate the interval of existence.

    2. Relevant equations

    y'=e^(x+y), i.v.p:y(0)=0

    3. The attempt at a solution

    this is my attempt:

    dy/dx=e^x+y=(e^x)(e^y)

    --> dy/e^y=(e^x)dx

    Integrating, -e^-y=e^x+C (C is constant) --> e^y=-e^x-C

    --> ln(e^-y)=ln(-e^x-C) --> y=-ln(-e^x-C)

    Because we have y(0)=0, 0=-ln(-1-C), so C=-2
    Therefore, y(x)=-ln(2-e^x) (=ln(1/(2-e^x)))
    Then, the interval of existence is (0, ln2).

    This is what i did, but I'm not confident for my work. So I want someone to look at it and help me if you find any mistake. Thanks!
     
  2. jcsd
  3. Aug 6, 2011 #2

    SammyS

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    Of course, x can be zero. y(0)= 0 . Right?

    In fact what makes you say that x can't be negative?
     
  4. Aug 6, 2011 #3
    you're right actually...

    so is the interval of existence actually (-infinity, ln2)?
     
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