Find the particular solution of the differential equation (tan⁻¹y - x)dy = (1 + y²)dx, given that when x = 0, y = 0.

Given:
(tan⁻¹y - x)dy = (1 + y²)dx
tan⁻¹y - x = (1 + y²)dx/dy
dx/dy + (1/(1+y²))x = tan⁻¹y/(1+y²)
The above equation is in the form of dx/dy + P(y)x + Q(y)
Here, P(y) = 1/(1+y²), Q(y) = tan⁻¹y/(1+y²)
Integrating factor = e∫1/(1+y²)dy = e^(tan⁻¹y)
So, xe^(tan⁻¹y) = ∫e^(tan⁻¹y)(tan⁻¹y/(1+y²))dy
Put tan⁻¹y = t => dy/(1+y²) = dt
xe^(tan⁻¹y) = ∫te^t dt
=> xe^(tan⁻¹y) = te^t - ∫e^t dt
xe^(tan⁻¹y) = te^t - e^t + C
=> xe^(tan⁻¹y) = (tan⁻¹y)e^(tan⁻¹y) - e^(tan⁻¹y) + C
Given that x = 0, y = 0, we get
0.e⁰ = (tan⁻¹0)e⁰ - e⁰ + C
0 = 0 - 1 + C
C = 1
So, the required solution is:
x = tan⁻¹y - 1 + e^(-tan⁻¹y)