Answer to Question #223874 in Chemical Engineering for Lokika

Question

Find the equation of the family of all erhogonal trajectories of the family ofcircles which pass through origin and hav centres on the y-axis.

Answer 1

Upon differentiation with respect to x, we obtain

2x+2yy′+2g(1−y′)=0,

where y′ denotes the derivative of y with respect to x. Upon dividing out by 2, we arrive at

x+yy′+g(1−y′)=0,

from which we get

g=x+yy′y′−1.

Putting this value of g into the equation of the family of circles, we get

x2+y2+2x+yy′y′−1(x−y)−2=0,

so

(x2+y2−2)(y′−1)+2(x+yy′)(x−y)=0

or

(x2+y2−2+2xy−2y2)y′+(2x2−2xy−x2−y2+2)=0

or

(x2+2xy−y2−2)y′+(x2−2xy−y2+2)=0,

from which we get

y′=−$\dfrac{x²−2xy−y²+2}{x²+2xy−y²−2}$

Now for the orthogonal trajectories, we get

$y′=\dfrac{x²+2xy−y²−2}{x²−2xy−y²+2}$