Find the particular solution of the differential equation: xdydx - y + xsin(yx) = 0, given that when x = 2, y = π.

xdydx - y + xsin(yx) = 0
Dividing throughout by x, we have
dydx - y/x + sin(yx) = 0
Let y/x = t, ∴ dy = tdx + xdt
Substituting these, we can write
t + xdt/dx - t - sin t = 0
⇒ dt/sin t = dx/x
⇒ ∫cosec t dt = ∫dx/x
∫cosec t (cosec t + cot t)/(cosec t + cot t) dt = lnx + c
∴ -ln(cosec t + cot t) = lnx + c
⇒ -c = ln(x cosec(y/x) + x cot(y/x))
Substituting when x = 2, y = π, we have
-c = ln(2 × 1 + 0) = ln2
Thus, we can write the solution as
⇒ 1 = 2(x cosec(y/x) + x cot(y/x))