Differential Equation related to orthogonal trajectories

• September 26th 2011, 08:11 PM
process91
[SOLVED] Differential Equation related to orthogonal trajectories
Find the orthogonal trajectories of all circles through the points (1,1) and (-1,-1)

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

$y'=\frac{y^2-2xy-x^2+2}{y^2+2xy-x^2-2}$

The equation for all circles which pass through the given points is:
$(x+C)^2+(y-C)^2=2+2C^2$
Solving for C:
$C = \frac{-x^2-y^2+2}{2 (x-y)}$
Implicit Differentiation of the original equation:
$2(x+C) + 2(y-C)y'=0$
$y'=\frac{x+C}{y-C}$
So, substituting C and taking the negative reciprocal, orthogonal trajectories must satisfy:
$y'=\frac{y^2-2xy-x^2+2}{y^2+2xy-x^2-2}$

And now I'm stuck. The answer in the book is the solution to this differential equation, but I have no idea how to get there from here aside from guessing the answer out of thin air.
• September 27th 2011, 07:13 AM
Jester
Re: Differential Equation related to orthogonal trajectories
You'll notice that your DE is

$\dfrac{dy}{dx} = \dfrac{(x-y)^2-2x^2+2}{(x+y)^2-2x^2-2}$

If we let $u = x+y$ and $v = x-y$ then the DE becomes

$\dfrac{dv}{du} = - \dfrac{u^2-v^2-4}{2uv}$

or

$2v\dfrac{dv}{du} = - \dfrac{u^2-v^2-4}{u}$

Letting $w = v^2$ now gives a linear ODE.
• September 27th 2011, 07:51 AM
process91
Re: Differential Equation related to orthogonal trajectories
Thank you very very much.

I'm working on Apostol Vol I. and this method was not covered at all. Is there any trick to getting the right substitution, or is it just a guess?
• September 28th 2011, 05:00 AM
Jester
Re: Differential Equation related to orthogonal trajectories
Sometimes it's just trial and error.
• September 28th 2011, 06:05 AM
Ackbeet
Re: Differential Equation related to orthogonal trajectories
Just to put my two cents' in: I'm not sure I could have easily reproduced what Danny did here. However, looking back on it, I would make the following claims.

1. Danny exploited the symmetries of the DE for all they were worth. You can see that both numerator and denominator have terms that look like a perfect square. So, he made a perfect square, and then thought of ways he could use that. (I should mention here that I am speculating as to what Danny was thinking - I'm really trying to reconstruct Danny's thoughts in my own head, which is always a perilous undertaking.)

2. Danny kept at it. You have to keep trying things. If at first something doesn't work, and you're stumped, then stop. Move on to something else, and then come back to it. Maybe that something else will even give you an idea on the first problem!

3. The above two points are what I would call "enabling the light-bulb". Often, in solving DE's, it is just a light-bulb moment that you need. You have to just "see the solution". If you follow 1 and 2 above, I think you'd put yourself in a better position to be able to "see it".

Hope that helps.