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2 problems

  1. Feb 25, 2006 #1
    Hi ,
    I have two problem with which I am stuck:

    1- Given the family of curves y=1/(x+C).
    Find the family of orthogonal trajectories

    For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
    What can I do here?

    2- The pairs {x^2+x,x^2} and {x,x+2x^2} ( found by the Wronskian) are base solutions of the equation y''+p(x)y'+q(x)y=0.
    Give the general solution of the equation.

    Here I dont even know how to start.

    Thank you
  2. jcsd
  3. Feb 25, 2006 #2
    Using 'k' instead of 'C', (in other words, let C=k)
    Let g(x) describe the family of orthagonal trajectories of y=f(x)=1/(x+k)

    y = f\left( x \right) = \frac{1}{{x + k}} \Rightarrow \frac{{df}}{{dx}} = - \frac{1}{{\left( {x + k} \right)^2 }} \Rightarrow \hfill \\
    \frac{{dg}}{{dx}} = \left( {x + k} \right)^2 \Rightarrow g\left( x \right) = \int {\left( {x + k} \right)^2 dx} = \frac{{\left( {x + k}\right)^3 }}{3} \hfill \\ \end{gathered} [/tex]
    Last edited: Feb 25, 2006
  4. Feb 26, 2006 #3


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    Why do say that's impossible? It's certainly possible for y' to be negative isn't it?
    It might be simpler, however, to reverse it: x+ C= 1/y so, differentiating both sides, 1= (-1/y2)y' and y'= -y2. For the orthogonal trajectories you must have y'= 1/y2.

    I'm not certain what is meant by "base solutions" or in what sense those are "base solutions" but I suspect that the general solution is
    y(x)= Cx2+ Dx.
  5. Feb 26, 2006 #4
    Thank you,
    I was confused by the fact that a square ca not be negative. But as you pointed out , we are talking about derivative..
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