Solve (1+y²)dx = (tan⁻¹y − x)dy

The equation is linear in x when y is the independent variable.

Rewrite as a linear equation in x

dx/dy = (tan⁻¹y − x)/(1 + y²) → dx/dy + x/(1 + y²) = tan⁻¹y/(1 + y²)

This is linear in x: dx/dy + Px = Q with P = 1/(1 + y²), Q = tan⁻¹y/(1 + y²).

Integrating factor

IF = e^(∫P dy) = e^(tan⁻¹y)

Solution

x e^(tan⁻¹y) = ∫ [tan⁻¹y/(1 + y²)] e^(tan⁻¹y) dy

Put t = tan⁻¹y, dt = dy/(1 + y²):

∫ t e^t dt = t e^t − e^t + C = (t − 1)e^t + C

So x e^(tan⁻¹y) = (tan⁻¹y − 1)e^(tan⁻¹y) + C

x = tan⁻¹y − 1 + C e^(−tan⁻¹y)