Question

Solve the differential equation:

`x sin (y/x) dy/dx + x - y sin (y/x) = 0`

Given that x = 1 when y = `π/2`.

Answer 1

We have,

`x sin (y/x) dy/dx + x - y sin (y/x) = 0`

⇒ `dy/dx = (y sin (y/x) - x)/(x sin (y/x))`   ...(i)

Above differential equation is a homogeneous equation

Put y = vx

Then, `dy/dx = v + x (dv)/dx`   ...(ii)

From (i) and (ii)

⇒ `v + x (dv)/dx = (vx.sin ((vx)/x) - x)/(x sin((vx)/x)`

⇒ `v + x (dv)/dx = (x(v sin v - 1))/(x sin v)`

⇒ `v + x (dv)/dx = (v sin v - 1)/sin v`

⇒ `x (dv)/dx = (v sin v - 1 - v sin v)/sin v`

⇒ `x (dv)/dx = - 1/sin v`

⇒ `sin v  dv = - 1/x dx`   ...[Here x ≠ 0]

Now, integrating both sides

⇒ `int sin v  dv = -int 1/x dx`

⇒ `- cos v = - log |x| + "C"`

Put `v = y/x`

⇒ `-cos(y/x) = - log |x| + "C"`   ...(iii)

Also, given that x = 1, when y = `π/2`

Put x = 1 and y = `π/2` in (iii)

⇒ `-cos(π/2) = -log 1 + "C"`

C = 0

⇒ `-cos(y/x) + log |x| = 0`

Therefore, `log |x| = cos(y/x)` is the required solution.