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Orthogonal Trajectory

  1. Oct 14, 2006 #1
    I have to find the orthogonal trajectory to the family of circles tangents to the x-axis.

    I eventually found that the derivative of the orthogonal trajectory would be:
    [tex]\frac{dy}{dx}= \frac{-x^2 + y^2}{2xy} [/tex]

    But how do i find the equation for "y" explicitly?

    (ps. This is not an assignment problem. I'm simply solving problems to get ready for my test)
    ps#2: I recently discovered that this may be a solvable differential equation. Perhaps Bernoulli?
  2. jcsd
  3. Oct 14, 2006 #2
    [tex] \frac{dy}{dx} + \frac{x^{2}+y^{2}}{2xy} = 0 [/tex]

    So [tex] \frac{dy}{dx} + \frac{x^{2}}{2xy} + \frac{y^{2}}{2xy} = 0 [/tex]

    [tex] \frac{dy}{dx} + \frac{x}{2y} = -\frac{y}{2x} [/tex] which is the form: [tex] \frac{dy}{dx} + P(x)y = Q(x)y [/tex] (i.e. the Bernouli equation is linear)
    Last edited: Oct 14, 2006
  4. Oct 15, 2006 #3


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    Do you mean with center on the y-axis also? You need another condition to have a one parameter family.

    Yes, that is correct for the orthogonal trajectories to the family of circles with center on the y-axis tangent to the x-axis.

    This is a "homogeneous" equation (note that "homogeneous" for first order equations is not the same as "homogeneous" for linear equations!): If you divide both numerator and denominator by x2 you get
    [tex]\frac{dy}{dx}= \frac{\frac{y^2}{x^2}- 1}{2\frac{y}{x}}[/itex]
    Now let v= y/x so that y= xv and y'= xv'+ v. Your equation becomes
    [tex]x\frac{dv}{dx}+ v= \frac{v^2-1}{v^2}= 1- \frac{1}{v^2}[/tex] That's a separable equation.

    (Courtigrad's suggestion that this is linear is completely wrong: 1/y is not linear.)
  5. Oct 15, 2006 #4
    was I correct on what I did before that then? It is Bernoulli right? And its courtrigrad, not courtigrad.
    Last edited: Oct 15, 2006
  6. Oct 15, 2006 #5
    made a mistake. should be:

    [tex] \frac{dy}{dx} - \frac{x^{2}-y^{2}}{2xy} = 0 [/tex]
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