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Orthogonal trajectory

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    i want to get the orthogonal trajectory of the curves of this family

    [tex] x^2 + y^2=cx [/tex]


    2. Relevant equations

    answer is given as : [tex] y^2 + x^2=cy [/tex]



    3. The attempt at a solution
    [tex] 2x + 2yy' = \frac {x^2 +y^2} {x} [/tex] then [tex] y' = \frac{y} {2x} - \frac{x}{y} [/tex]
    let v=y/x ;
    [tex] x\frac{dv}{dx} =\frac {-1}{2v} [/tex]
    thus:
    [tex] - v^2 = \ln |x| [/tex] or
    [tex] xe^{\frac {y^2}{x^2} } = c [/tex]
    which is far from the given answer . Got problem from odinary differential equations by rahman volume 1, 1994,
     
  2. jcsd
  3. Sep 21, 2009 #2
    Should be [tex] y' = \frac{y} {2x} - \frac{x}{2y} [/tex]

    I didn't get anything close to that, even using the incorrect y'.

    [tex] y' = \frac{y} {2x} - \frac{x}{2y} =\frac{y^2-x^2}{2xy} [/tex]

    so the orthogonal trajectory satisfies

    [tex] y' = -\frac{2xy}{y^2-x^2} [/tex]

    Then use v=y/x as you suggested
     
  4. Sep 22, 2009 #3
    thanks for the corrections , i made errors in my working , so following from where u left i get stuck here; [tex] \frac{(v^2 -1)dv} {-v(v^2 + 1)} = \frac {dx}{x} [/tex]
     
  5. Sep 22, 2009 #4
    Thanks was able to integrate where billy bob left of using integrating factor[tex] \frac {1}{y^2} [/tex]
    for anyone who gets stranded
     
  6. Sep 22, 2009 #5
    Here's how I got past that step you mentioned:

    [tex]-\frac{v^2-1}{v^3+v}=\frac{1}{v}-\frac{2v}{v^2+1}[/tex]

    Glad it worked out for you.
     
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