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Diff.Eq. Seperation of variables.

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the given differential equation subject to the indicated initial condition.

    (e^-y + 1)sinxdx=(1+cosx)d, y(0)=0


    2. Relevant equations
    Basically we have to use seperation of varaibles to solve before using initial value condition.


    3. The attempt at a solution
    After separation of variables

    Dy/(e^-y +1) = sinx dx/(1+cosx)

    take the integral of both sides
    ∫Dy/(e^-y +1)=ln|e^-y+1|+y

    ∫sinx dx/(1+cosx)= -ln(1+cosx)+c

    Clean it up a bit
    ln|e^-y+1|+y= -ln(1+cosx)+c


    I have no idea what to do now with the y(0)=0
    That means plus in x=0 for the equation correct?
     
    Last edited: Feb 25, 2014
  2. jcsd
  3. Feb 25, 2014 #2

    SteamKing

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    You use the initial condition to determine the value of c.

    when x = 0, y = 0.
     
  4. Feb 25, 2014 #3
    How did you get y=0?
     
  5. Feb 25, 2014 #4

    pasmith

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    That's what [itex]y(0) = 0[/itex] means!
     
  6. Feb 25, 2014 #5

    ChrisVer

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    well still there is a problem if x(0) is not zero for y(0)=0...
    In that case you solve c with respect to x(0) (a parameter) and input it
     
  7. Feb 25, 2014 #6
    okay thank you.
    My last question is.
    can someone please check my integration
    ∫Dy/(e^-y +1)=ln|e^-y+1|+y

    to make sure ^ is correct?
     
  8. Feb 25, 2014 #7

    ChrisVer

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    you can always try to take the derivative of the righthand side and check by yourself ;) much easier
     
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