Solve the differential equation: (tan⁻¹y - x)dy = (1 + y²)dx

On rearranging the equation becomes:
(tan⁻¹y - x)dy = (1 + y²)dx
⇒dx/dy + x/(1 + y²) = tan⁻¹y/(1 + y²).
This is in the form dx/dy + P(y).x = Q(y), so it is a non-homogeneous linear differential equation of the first order.
The integrating factor is:
I.F. = e∫dy/(1+y²) = e^(tan⁻¹y)
Multiplying both sides of the equation with integrating factor, we get:
e^(tan⁻¹y)(dx/dy + x/(1 + y²)) = e^(tan⁻¹y)tan⁻¹y/(1 + y²)
⇒e^(tan⁻¹y)dx/dy + e^(tan⁻¹y)x/(1 + y²) = e^(tan⁻¹y)tan⁻¹y/(1 + y²)
⇒d/dy(xe^(tan⁻¹y)) = e^(tan⁻¹y)tan⁻¹y/(1 + y²)
Integrating by parts, by taking tan⁻¹y as first function and e^(tan⁻¹y)/(1 + y²) as second function we get:
⇒xe^(tan⁻¹y) = ∫e^(tan⁻¹y)tan⁻¹y/(1 + y²)dy
= tan⁻¹y∫e^(tan⁻¹y)/(1 + y²)dy - ∫d(tan⁻¹y)(∫e^(tan⁻¹y)/(1 + y²)dy)
= tan⁻¹y.e^(tan⁻¹y) - ∫dy/(1 + y²)(∫e^(tan⁻¹y)/(1 + y²)dy)
⇒xe^(tan⁻¹y) = tan⁻¹y.e^(tan⁻¹y) - e^(tan⁻¹y) + C
⇒x = tan⁻¹y + Ce^(-tan⁻¹y)