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Homework Help: Differential Equation Help

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    i have the equation dy/dx = [tex]\frac{(y/x)-4}{1-(y/x)}[/tex]

    i am told that v(x)=xy

    and v=y/x

    Express this using v, dv/dx, and x


    I can get dy/dx = [tex]\frac{(v)-4}{1-(v)}[/tex] but that's it

    2. Relevant equations

    ??

    3. The attempt at a solution

    I don't even know how to do dv/dx

    can anyone help. im so confused?
     
  2. jcsd
  3. Jan 20, 2009 #2

    rock.freak667

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    If y=Vx, then dy/dx = V+x(dV/dx)


    [tex]V + x \frac{dV}{dx}= \frac{V-4}{1-V}[/tex]

    make dV/dx the subject it now.
     
  4. Jan 20, 2009 #3

    quasar987

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    Which of the two is it??

    You are told that y is a function of x and that its derivative dy/dx satisfies

    [tex]\frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}[/tex]

    But this differential equation is complicated. We would hope to simplify it by considering a new function v(x) that we would define as v(x)=y(x)/x so that the right hand side of the differential equation becomes

    [tex]\frac{v-4}{1-v}[/tex]

    like you said. So it is v=y/x that is useful for simplifying the equation, not v=yx.

    To find dv/dx, use the chain rule. rock.freak667 gave you the answer but make sure you know how to obtain it yourself.
     
  5. Jan 21, 2009 #4
    i don't understand how the chain rule applies in this situation!!

    help!! how do i take the derivative of (v-4)/(1-v) with respect to x?? there is no x!!!
     
  6. Jan 21, 2009 #5
    You are using implicit differentiation to get to what rock said, if you see how he got that, then just solve for dV/dx, at that point it's just a separable DE
     
  7. Jan 21, 2009 #6

    quasar987

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    Sorry, it's not the chain rule, it's just the rule for differentiating a product of function. Anyway, you have expressed the right hand side of

    [tex]
    \frac{dy}{dx}=\frac{(y/x)-4}{1-(y/x)}
    [/tex]

    as a function of v, which you have defined as v=y/x. Now you'd like to express the left hand side as a function of v also. So you compute

    [tex]\frac{dv}{dx}=\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}[/tex]

    Solving for dy/dx gives you what rockfreak wrote and now you've got a simpler differential equation in v as a function of x to solve:

    [tex]
    v + x \frac{dv}{dx}= \frac{v-4}{1-v}
    [/tex]
     
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