Question

Show that the given differential equation is homogeneous and solve them.

`{xcos(y/x) + ysin(y/x)}ydx = {ysin (y/x) -  xcos(y/x)}xdy`

Answer 1

Given differential equation

`{x cos (y/x) + y sin (y/x)} y dx = {y sin (y/x) - x cos (y/x)} x dy`

or `dy/dx = ({x cos (y/x) + y sin (y/x)} y)/({y sin (y/x) - x cos (y/x)} x)`

and `dy/dx = ((y/x) {cos (y/x) + y/x sin (y/x)})/({y/x sin (y/x) - cos (y/x)} x) = g (y/x)` (say)             .... (i)

The right side of the differential equation is in the form of `g (y/x)`. Therefore, this is an even exponential differential equation of zero degree.

∴ Putting y = vx 

v + x `(dv)/dx = ((cos v + v sin v) v)/(v sin v - cos v)`

⇒ x `(dv)/dx = (v cos v + v^2 sin v)/(v sin v - cos v) - v`

= v cos v + v² sin v

⇒ x `(dv)/dx = (- v^2 sin v + v cos v)/(v sin v - cos v)`

⇒ x `(dv)/dx = (2v cos v)/(v sin v - cos v)`

`= (v sin v - cos v)/(v cos v)  dv = 2/x  dx`

`= (tan v - 1/v) dv = 1/x  dx`

On integrating

log sec v - log v = 2 log x + log C

log `((sec v)/v)` = log x² = log C

log `((sec v)/v)` = log cx²

sec v = v. Cx²

Finally, on putting `y/x` in place of v

`sec (y/x) = (y/x). Cx^2`

`sec (y/x) = Cxy`

`xy  cos |y/x| = C`