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Separation of variables question

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the following separable equation

    [itex]\frac{dy}{dx} = \frac{y}{x(x-1)}[/itex]

    2. Relevant equations

    [itex]\int\frac{1}{x}dx=ln(x)[/itex]


    3. The attempt at a solution

    See attachment

    Im getting [itex]y=(\frac{x-1}{x})^{c}[/itex] as my answer when in fact the answer the tutor gave is [itex]y=c(\frac{x-1}{x})[/itex]

    not sure what im missing
     

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  2. jcsd
  3. Sep 13, 2011 #2
    I don't understand why you say that:
    [itex]\int\frac{dx}{x(1-x)} = \int \frac{cdx}{1-x}- \int \frac{cdx}{x}[/itex]

    What is this c?
    You should be writing:
    [itex]\int\frac{dx}{x(1-x)} = \int (\frac{a}{1-x} + \frac{b}{x})dx[/itex]
    and find the appropriate a and b.
     
  4. Sep 13, 2011 #3
    c is a constant value

    a and b would be the same value, since the numerator on the [itex]\int\frac{dx}{x(1-x)} [/itex] is 1

    you just have to treat [itex] (\frac{a}{1-x} + \frac{b}{x}) [/itex] like adding fractions and you'll realise why I simply put c on the top of both fractions
     
  5. Sep 13, 2011 #4
    So if you know c = 1, why do you keep on dragging it?
    It's not the "c" in the final answer.
    The "c" in the final answer comes as a result of the integration constant.

    [itex]\int \frac{dx}{x(x-1)} = ln(\frac{x}{x-1}) + C[/itex]

    EDIT: I just reread your post - maybe you weren't aware that your c=1? Of course I realize that a and b have the same values (actually opposite values) - but you should also notice that a = -b = 1.
     
    Last edited: Sep 13, 2011
  6. Sep 13, 2011 #5

    lurflurf

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    Homework Helper

    You added a false constant and omitted the actual constant.

    dy/y=dx/[x(x-1)]
    look like log write as log
    d log(y)=d log(1-1/x)
    integrate
    y=C (1-1/x)

    1/[x(x-1)]=1/(x-1)-1/x=(1/x)[1/(1-1/x)-(1-1/x)/(1-1/x)]
    =(1/x^2)/(1-1/x)=(1-1/x)'/(1-1/x)=d log(1-1/x)
    or
    1/[x(x-1)]=1/(x-1)-1/x=d log(x-1)-d log(x)=d log(1-1/x)
     
  7. Sep 13, 2011 #6

    I was aware that c=1 from adding the fractions together

    I explored the idea of using integration constant, however I backed out because I wouldn't be able to find an inverse natural log of it.

    Im guessing cause C (integration constant ) is an arbitrary number, we can simply rewrite it as ln(C) ?
     
  8. Sep 14, 2011 #7
    You could do that, or you could look at it like this:

    [itex]ln(y) = ln(\frac{x}{x-1}) + c => y(x) = e^{ln(\frac{x}{x-1}) + c} =>

    y(x) = e^{c}ln(\frac{x}{x-1}) => y(x) = Cln(\frac{x-1}{x})^{-1} =>
    y(x) = -Cln(\frac{x-1}{x}) =>
    y(x) = Cln(\frac{x-1}{x})[/itex]

    The example shows how C can alter it's value (c --> ec --> -ec). However I didn't bother to "rename" it every time, because it's clear that it's just a constant.

    You'll always have C popping in a first order PDE, and that's because you'll always have a constant of integration. That's always the source of c.
     
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