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Differential Equation

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a solution of [tex]x\frac{dy}{dx} = y^{2} - y[/tex] that passes through the indicated points.

    a) (0,1)
    b) (0,0)
    c) (1/2 , 1/2)
    d) (2, 1/4)

    2. Relevant equations

    [tex]x\frac{dy}{dx} = y^{2} - y[/tex]

    3. The attempt at a solution

    [tex]x\frac{dy}{dx} = y^{2} - y[/tex]

    [tex]\frac{dy}{y^{2} - y} = \frac{dx}{x}[/tex]

    [tex]\int\frac{dy}{y^{2} - y} = \int\frac{dx}{x}[/tex]

    [tex]\int\frac{dy}{y(y - 1)} = ln|x| + c[/tex]

    I used partial fractions to solve the left:

    [tex]\int\left(\frac{1}{y - 1} - \frac{1}{y}\right)dy = ln|x| + c[/tex]

    [tex]\int\frac{1}{y - 1}dy - \int\frac{1}{y}dy = ln|x| + c[/tex]

    [tex]\int\frac{1}{y - 1}dy - ln|y| = ln|x| + c[/tex]

    u = y - 1
    du = dy

    [tex]\int\frac{1}{u}du - ln|y| = ln|x| + c[/tex]

    [tex]ln|u| - ln|y| = ln|x| + c[/tex]

    [tex]ln|y - 1| - ln|y| = ln|x| + c[/tex]

    [tex]ln|\frac{y - 1}{y}| = ln|x| + c[/tex]

    [tex]\frac{y - 1}{y} = e^{ln|x| + c} = (e^{ln|x|})(e^{c})[/tex]

    [tex]\frac{y - 1}{y} = |x|(e^{c})[/tex]

    [tex]\frac{y - 1}{y} = \pm(e^{c})(x)[/tex]

    [tex]\frac{y - 1}{y} = c(x)[/tex]

    Now this is where I am confused:

    a) (0, 1)

    [tex]\frac{1 - 1}{1} = c(0)[/tex]

    [tex]0 = 0[/tex]

    Does this mean c = 0 or does it mean that there is no solution through the point (0, 1) or that there is a solution through the point as left = right?
     
  2. jcsd
  3. Sep 30, 2009 #2

    tiny-tim

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    Hi KillerZ! :smile:
    Nooo :redface:

    ln|(y-1)/xy| = c,

    so … ? :smile:
     
  4. Sep 30, 2009 #3
    I do not understand where the xy in ln|(y-1)/xy| came from.
     
  5. Sep 30, 2009 #4

    tiny-tim

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    lnA - lnB = ln(A/B) :wink:
     
  6. Sep 30, 2009 #5
    So.

    [tex]ln|y - 1| - ln|y| = ln|x| + c[/tex]

    [tex]ln|y - 1| - ln|y| - ln|x| = c[/tex]

    becomes ?

    [tex]ln|\frac{y - 1}{yx}| = c[/tex]
     
  7. Sep 30, 2009 #6

    tiny-tim

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    :tongue2: go on … :smile:
     
  8. Sep 30, 2009 #7
    [tex]ln|\frac{y - 1}{yx}| = c[/tex]

    [tex]\frac{y - 1}{yx} = e^{c}[/tex]

    replace e^c with c as it is just a constant

    [tex]\frac{y - 1}{yx} = c[/tex]
     
  9. Sep 30, 2009 #8

    tiny-tim

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    ah, looking back, I see you got there anyway …

    ok, you were puzzled about getting it to go through (0,1).

    You have c|x| = |(y-1)/y|, so for any non-zero value of c, limy->1|x| = 0 …

    they all go through (0,1) ! :smile:
     
  10. Sep 30, 2009 #9
    ok thank you.
     
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