# Orthogonal trajectories

1. Sep 16, 2009

### intenzxboi

The set of orthogonal trajectories for the family indicated by
( x-c)^2 + y^2 = c^2

My work:

y' = -(x-c)/y Since c= ( x^2 + y^2 ) / 2x

plugging back in and doing -1/y' i got

y' = 2xy / ( x^2 - y^2)

Then im supposed to move the x and y to a side and integrate but i don't see how it is possible.

2. Sep 16, 2009

### HallsofIvy

You can't. This is NOT a "separable" differential equation. It is, however, a "homogeneous" equation because replacing x by a x and y by ay on the right gives 2(ax)(ay)((ax)^2- (ay)^2)= 2a^2xy/(a^2(x^- y^2)= 2xy/(x^2- y^2) again. That essentially means that the right hand side can be written as a function of y/x. Let v= y/x so that y= xv, y'= xv'+ v and replace y in the right side by xv to get a separable equation for v as a function of x.