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Old exam question

  1. Nov 8, 2005 #1
    Could someone please just give me a hint to get started.
    [tex]\frac{dy}{dx}-\frac{y}{x}=\frac{y^2}{x^2} for x>0[/tex]
    thanks
    Skook
     
  2. jcsd
  3. Nov 8, 2005 #2
    [tex]
    \frac{{dy}}{{dx}} - \frac{y}{x} = \frac{{y^2 }}{{x^2 }} \Rightarrow \frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right) + \left( {\frac{y}{x}} \right)^2
    [/tex]

    Let y = v(x)x. Is this a standard substitution for the subject you are studying?
     
  4. Nov 9, 2005 #3
    thanks for that

    I hope the solution is [tex]y=-\frac{x}{\ln{Cx}}[/tex]. It was from an Open University course (http://www3.open.ac.uk/courses/bin/p12.dll?C02MS324) [Broken]. First maths course I've done in over 25 years...............
     
    Last edited by a moderator: May 2, 2017
  5. Nov 9, 2005 #4
    :approve: I think so, too
     
  6. Nov 9, 2005 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since x and y only appear together as y/x, try the obvious substitution: Introduce a new dependent variable [itex]v= \frac{y}{x}[/itex].

    Then y= vx so [itex]\frac{dy}{dx}= x\frac{dv}{dx}+ v[/itex]

    Your equation becomes
    [tex]x\frac{dv}{dx}- v= v^2[/tex] or
    [tex]x\frac{dv}{dx}= v^2+ v[/tex]
    a separable equation.
     
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