Question

find the orthogonal trajectories for the family of curves e^-x cosy = C.

Answer 1

We have seen above the orthogonal trajectories of u = c1 are given by v = c2 where v is the harmonic conjugate of u.

:.u=e−xcosy+xy

Differentiating w.r.t x&y

ux=−e−xcosy+yanduy=−exsiny+x

Also f1(z)=ux+ivx=ux–iuy [By C-R equations]

=(−e−xcosy+y)–i(−e−xsiny+x)

By Milne – Thompson’s method , we replace x by z and y by zero.

:.f1(z)=−e−z–iz

By integration f(z)=e−z–iz22+c

:.f(z)=e−(x+iy)–i(x+iy)22+c=e−xe−iy−i2(x2+2ixy–y2)+c=e−x(cosy–isiny)−i2(x2–y2)+xy

Now only considering imaginary part

v=e−xsiny−12(x2–y2)

:. The required orthogonal trajectories are

e−xsiny+12(x2–y2)=c.