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How can I find the orthogonal trajectories of the curves 1) $y^{2}=cx, x^{2}y=c $

and

2) $y=c\sin x$

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The key fact is that solutions of ${dy\over dx}=f(x,y)$ will be orthogonal to solutions of ${dy\over dx}=-{1\over f(x,y)}$.

Thus, $y^2=c x\implies 2y{dy\over dx}=c\implies {dy\over dx}={c\over 2y}$. Solving for $c$ in the original equation, $c=y^2/x$. Substituting this into the derivative to eliminate $c$, we obtain ${dy\over dx}={y\over 2x}$. Thus, from the key fact above, the orthogonal trajectories obey ${dy\over dx}={-2x\over y}$. Solve this separable ODE to obtain the implicit family of curves $y^{2}=-2x^2+2K$.

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A similar process will get the other orthogonal trajectories.

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  • $\begingroup$ So,at the subquestion 1,do I not have to use the equation $x^{2}y=c$ ?? $\endgroup$
    – evinda
    Nov 28 '13 at 18:11
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    $\begingroup$ Perhaps they want you to find the orthogonal trajectories for the first family and then the second family? $\endgroup$
    – JohnD
    Nov 28 '13 at 18:18
  • $\begingroup$ For the second family,I found $y=\pm \sqrt{x^{2}+c}$ ..Am I right???And how can I continue now?? $\endgroup$
    – evinda
    Nov 28 '13 at 18:29
  • $\begingroup$ Do you understand the first sentence in my answer? If not, concentrate there. $\endgroup$
    – JohnD
    Nov 28 '13 at 18:32
  • $\begingroup$ I don't really understand :/ Could you explain it further to me?? $\endgroup$
    – evinda
    Nov 28 '13 at 18:37

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