Show that the differential equation [xsin2(yx) - y]dx + xdy = 0 is homogeneous. Find the particular solution of this differential equation, given that y = π/4 when x = 1.

We have [xsin2(yx) - y]dx + xdy = 0
dy/dx = -x/[xsin2(yx) - y] = f(x,y) — (1)
Put x = kx, y = ky in equation (1),
f(kx, ky) = (-kx/[kxsin2(ky/kx) - ky]) = k^0(-x/[xsin2(y/x) - y]) = k^0f(x,y)
So it is clear that the given equation is a homogenous differential equation of degree 0.
Now, let y = vx in (1), then dy/dx = v + xdv/dx
Therefore, v + xdv/dx = -x/[xsin2(v) - vx]
=> v + xdv/dx = -sin2v + v
=> -∫cosec2v dv = ∫dx/x
=> cotv = log|x| + c
=> cot(y/x) = log|x| + C
Since given that y = π/4 when x = 1, so, cot(π/4) = log|1| + C
=> c = 1
The required solution is cot(y/x) = log|x| + 1