Find the particular solution of the following differential equation: xydy/dx = (x+2)(y+2); y = -1 when x = 1

Given differential equation is
xydy/dx = (x+2)(y+2)
(y/(y+2))dy = ((x+2)/x)dx
Integrating both sides,
∫(y/(y+2))dy = ∫((x+2)/x)dx
∫(1 - 2/(y+2))dy = ∫(1 + 2/x)dx
y - 2log|y+2| = x + 2log|x| + C
Given that y = -1 when x = 1
Therefore, -1 - 2log|1| = 1 + 2log|1| + C
⇒C = -2
Therefore the required solution is
y - 2log|y+2| = x + 2log|x| - 2