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Algebra of Differential Equation

  1. Nov 16, 2009 #1
    I need to solve dy/dx=(y-1)^2
    dy/(y-1)^2=dx
    dy/u^2=x+c
    -u^-1=x+c
    -1/y-1=x+c
    -1=(x+c)(y-1)
    y-1=-1/(x+c)
    y=(x+c-1)/(x+c)
    y(0)=1
    1=c-1/c
    1=1-1/c
    I don't know where my algebra is messing up?
     
  2. jcsd
  3. Nov 16, 2009 #2

    Mark44

    Staff: Mentor

    Are you sure about your initial condition y(0) = 1? My work is the same as yours. If you look at the graph of y = 1 - 1/(x + c), there is a vertical asymptote at x = -c, and a horizontal asymptote for the line y = 1. No matter what c is, y will never equal 1.
     
  4. Nov 17, 2009 #3
    Yes, the initial condition is y(0)=1. I have looked in the back of the book and the answer is in fact y=1.
     
  5. Nov 17, 2009 #4

    Mark44

    Staff: Mentor

    Yep, that works. To take this into account, we need to look at this step:
    dy/(y - 1)^2 = dx
    The left side is defined only if y is not 1. If y = 1, the differential equation is y' = 0, so y = K. With your initial condition, this gives y(x) = 1.

    For any other initial condition, y(x) is as you had it.
     
  6. Nov 18, 2009 #5
    Ok, I guess that makes sense, but I was hoping I could solve it algebraically.
     
  7. Nov 18, 2009 #6

    Mark44

    Staff: Mentor

    Isn't y' = 0 ==> y = K algebraic enough for you? I wouldn't describe this as an "algebraic" solution, though. This is a very simple differential equation, and its solution is equally simple.
     
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