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Initial condition problem

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    They give me y ' = (xysinx)/ (y+1) , y(0) = 1

    2. Relevant equations

    So I just separated and integrated

    3. The attempt at a solution

    I'm 99 percent sure I'm OK up till here. I just wanted to get an explanation for something.

    I was wondering my answer is y + ln(y) = sinx -xcosx +C
    So there is no way to write this so I just have y on the right. So what am I to say? There is an example similar in my book they does this mean I can't find the constant?????
     
  2. jcsd
  3. Oct 26, 2013 #2
    Sure. This differential equation is variable separable.

    I think that the initial condition is, at x=0, y=1. Well the equation can be simplified as,

    ln(yey) = sinx -xcosx +C
    yey = ke(sinx -xcosx), where k is another constant. But does this lead to anything ?
     
  4. Oct 26, 2013 #3
    ln(y(e^y)) = sinx -xcosx +C
    How did you get this?
     
  5. Oct 26, 2013 #4

    LCKurtz

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    Just put x=0 and y=1 in that last equation to evaluate C.
     
  6. Oct 27, 2013 #5
    This last equation ?
    y + ln(y) = sinx -xcosx +C
     
  7. Oct 27, 2013 #6

    vela

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    Yes.
     
  8. Oct 27, 2013 #7
    Got it thanks
     
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