Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0

The equation is not linear in y, but it is linear in x taken as the dependent variable.

Rewrite as a linear equation in x

Multiply the equation by dx/dy (using (dy/dx)(dx/dy) = 1):

(1 + y²) dx/dy + x − e^(tan⁻¹y) = 0, so

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

This is linear in x with P = 1/(1 + y²) and Q = e^(tan⁻¹y)/(1 + y²).

Integrating factor

IF = e^(∫dy/(1+y²)) = e^(tan⁻¹y)

Solution

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

Put t = tan⁻¹y: ∫ e^(2t) dt = e^(2t)/2 + C.

x e^(tan⁻¹y) = e^(2 tan⁻¹y)/2 + C

x = ½ e^(tan⁻¹y) + C e^(−tan⁻¹y)

Check: put x = ½e^t + Ce^(−t), t = tan⁻¹y: dx/dy = (½e^t − Ce^(−t))/(1+y²) and dx/dy + x/(1+y²) = e^t/(1+y²) ✓.