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Ordinary differential equation:

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Obtain the general solution:
    [tex](1 - x)y' = y^2[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex](1 - x)\frac{dy}{dx} = y^2[/tex]

    [tex](1 - x)dy = y^2dx[/tex]

    [tex]\frac{dy}{y^2} = \frac{dx}{(1-x)}[/tex]
    integrating both sides:

    i used ln on the constant at the right side
    [tex]-\frac{1}{y} = \ln(1 - x) + \ln{c}[/tex]

    so my answer is:
    [tex]-1 = y\ln{(c(1 - x))}[/tex]

    the answer seems to be different at the back of my book w/c is
    [tex] 1 = y\ln{(c(1 - x))}[/tex] <<< no negative as before

    can you tell me what's wrong with my soln?
     
  2. jcsd
  3. Nov 14, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Here is the problem. What do you get if you differentiate ln(1-x)? (Don't forget to use the chain rule!)

     
  4. Nov 14, 2007 #3
    oh yeah! forgot about that thanks alot :D
     
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