Old exam question

  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]
  2. jcsd
  3. [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

    Let y = v(x)x. Is this a standard substitution for the subject you are studying?
  4. 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). First maths course I've done in over 25 years...............
  5. :approve: I think so, too
  6. HallsofIvy

    HallsofIvy 41,270
    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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