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Homework Help: Differential Equation related to orthogonal trajectories

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the orthogonal trajectories of the given family of curves:
    All circles through the points (1,1) and (-1,-1)

    I have reduced the problem to finding solutions to the following differential equation:


    2. Relevant equations

    3. The attempt at a solution
    I believe the equation for all circles which pass through the given points is:
    Solving for C:
    [tex]C = \frac{-x^2-y^2+2}{2 (x-y)}[/tex]
    Implicit Differentiation of the original equation:
    [tex]2(x+C) + 2(y-C)y'=0[/tex]
    So, substituting C and taking the negative reciprocal, orthogonal trajectories must satisfy:

    And now I'm stuck. The answer in the book agrees with the steps I've taken so far, but I have no idea how to get there from here aside from guessing the answer out of thin air.
  2. jcsd
  3. Sep 26, 2011 #2


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    Scratch that - read problem incorrectly
    Last edited: Sep 27, 2011
  4. Sep 26, 2011 #3
    Aren't the equations you mentioned circles which have their centers at (1,1) and (-1,-1) respectively? They don't go through those points (I know, "go through" is an ambiguous term at best, but here the author definitely means all circles which contain the points (1,1) and (-1,-1)).

    The answer is definitely the solution to that differential equation (I implicitly differentiated the answer given in the book and got my differential equation), but I just can't solve it.
  5. Sep 27, 2011 #4


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    The situation is similar to the other problem you had with a family of circles, those all intersecting at ( -1, 0 ) and ( 1, 0 ). I confirm your result for the circle equation; the graph for the family looks like the set from that other problem, rotated 45º and enlarged by a factor of √2 . I've already checked that the differential equation for the orthogonal set is again not exact. In view of the hour (for me), I'm going to intuit that you have a Bernoulli equation again, perhaps even with the same index n = -1 .
  6. Sep 27, 2011 #5
    Just wanted to post that I got help with this in another forum, the solution was to take u=x+y and v=y-x, then consider dv/du. This results in a solvable Bernoulli equation. For the record, Apostol does not cover anything even close to this sort of substitution.
  7. Sep 27, 2011 #6


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    That's similar to the idea I finally had this morning (I did say I'm usually little help after midnight...). Since we know how to solve the problem with the intersection points on the x-axis, make a rotational transformation of the plane. We will have ( -1, -1 ) end up at ( -√2 , 0 ) and ( 1, 1 ) go to ( √2 , 0 ) if we use new variables

    [itex]X = \frac{1}{\sqrt{2}} ( x + y ) [/itex] and [itex]Y = \frac{1}{\sqrt{2}} ( x - y ) [/itex].

    This should eliminate the troublesome xy-terms and produce a re-scaled version of the other problem you did. The method you got from the other folks does this rotation and expands the scale, but the transformed DE still looks similar to the one you did before.

    Apostol may be assuming that the student is already familiar with this type of transformation; this is frequently the case in "high-numbered" problems at the end of a section in a textbook. Authors will include such problems which draw on ideas from other branches of mathematics. (This problem is a good example of using the graph of each problem to get ideas for what to try.)

    I'm glad you were able to get help from someone.
  8. Sep 27, 2011 #7
    I think it's possible he was trying to draw on those other ideas, although one of the strengths of this book so far for me has been that it is fairly self-contained. I see that from the graph you could guess that the solution must be a bunch of circles, and get an idea on where their centers must lie. After a few tests of various equations and their derivatives, you are bound to come upon the correct answer, and this is very likely what he expected however I don't like guess/check methods like that.

    Thanks again for helping me.
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