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Differentials with linear equations

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that y(x) is the solution to the initial problem, y'=y(1-x), y(1)=e
    find y(2)


    2. Relevant equations



    3. The attempt at a solution
    This is my initial attempt:
    [itex]\frac{dy}{dx}[/itex]=y(1-x)
    [itex]\frac{dy}{y}[/itex]=(1-x)dx

    i then integrated both side to get:
    lny=-ln(1-x)+C

    and here's the problem, if i plug in 1 to find c, ln(1-x) does not exist at that point....
     
    Last edited: Mar 26, 2012
  2. jcsd
  3. Mar 26, 2012 #2
    I tried another approach:
    y'+yx=y
    (yx)'=y

    integrated both sides:
    yx=[itex]\frac{1}{2}[/itex]([itex]y^{2}[/itex]) +C

    but with this approach i wasn't able to solve for y
     
  4. Mar 26, 2012 #3
    First of all, you integrated the right hand side incorrectly. Remember, you're just integrating a polynomial.
     
  5. Mar 26, 2012 #4
    Wait a minute, (yx)'=xy'+yx'=xy'+y, not y'+yx.
     
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