Show that the differential equation (xex/y+y)dx=xdy is homogeneous. Find the particular solution of this differential equation, given that x=1 when y=1.

The given differential equation can be written as
dx/dy = x/(xex/y + y) — (1)
Let F(x,y) = x/(xex/y + y)
Then F(λx, λy) = λx/(λx e^(λx/λy) + λy) = λx/(λx e^(x/y) + λy) = x/(x e^(x/y) + y) = F(x,y)
Thus, F(x,y) is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation.
To solve it, we make the substitution x = vy — (2)
Differentiating equation (1) and (2) w.r.t y, we get
dx/dy = v + y dv/dy
Substitute the value of x and dx/dy in equation (1), we get
v + y dv/dy = v/(ve^v + 1)
y dv/dy = v/(ve^v + 1) - v
y dv/dy = (v - v^2e^v - v)/(ve^v + 1)
y dv/dy = -v^2e^v/(ve^v + 1)
(ve^v + 1)/v^2e^v dv = -dy/y
∫(ve^v + 1)/v^2e^v dv = -∫dy/y
∫(1/v + 1/v^2e^v)dv = -log|y| + C
∫(1/v)dv - ∫(e^-v/v)dv = -log|y| + C
log|v| - Ei(-v) = -log|y| + C
Replacing v by x/y
log|x/y| - Ei(-x/y) = -log|y| + C — (3)
Substituting x = 1 and y = 1 in equation 3, we get
log|1/1| - Ei(-1) = -log|1| + C
0 - Ei(-1) = 0 + C
=> C ≈ 0.2194
Therefore, the solution is
log|x/y| - Ei(-x/y) = -log|y| + 0.2194