Find the particular solution of the differential equation dx/dy + x cot y = 2y + y² cot y, (y ≠ 0), given that x = 0 when y = π/2.

We have dx/dy + x cot y = 2y + y² cot y, (y ≠ 0)
It is clear that this is a linear differential equation of the form dx/dy + P(y)x = Q(y)
Here, P(y) = cot y, Q(y) = 2y + y² cot y
Integrating factor = e∫cot y dy = elog sin y = sin y
So, solution is given by, x sin y = ∫sin y (2y + y² cot y) dy
x sin y = ∫2y sin y dy + ∫y² cos y dy
x sin y = ∫2y sin y dy + y²∫cos y dy - ∫(d(y²)/dy ∫cos y dy) dy
x sin y = ∫2y sin y dy + y² sin y - ∫2y sin y dy
x sin y = y² sin y + C
Given that x = 0, when y = π/2, we get
0 sin(π/2) = (π/2)² sin(π/2) + c
Therefore, c = -π²/4
So, the required solution is:
x sin y = y² sin y - π²/4