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Solve (1+x²)dy/dx+y=e^(tan⁻¹x)

Short answer

Rewrite as linear in y with integrating factor e^{tan⁻¹x}. Multiply, substitute t=tan⁻¹x, and integrate to obtain y = (1/2)e^{tan⁻¹x} + c e^{-tan⁻¹x}.

Solve the differential equation (1 + x²) dy/dx + y = e^(tan⁻¹x).

(1 + x²) dy/dx + y = e^(tan⁻¹x)

Standard linear form

dy/dx + y/(1 + x²) = e^(tan⁻¹x)/(1 + x²)

P = 1/(1 + x²), Q = e^(tan⁻¹x)/(1 + x²).

Integrating factor

IF = e^(∫dx/(1 + x²)) = e^(tan⁻¹x)

Solution

y e^(tan⁻¹x) = ∫ [e^(tan⁻¹x)/(1 + x²)] e^(tan⁻¹x) dx

Put t = tan⁻¹x, dt = dx/(1 + x²):

∫ e^(2t) dt = e^(2t)/2 + c

y e^(tan⁻¹x) = ½ e^(2 tan⁻¹x) + c

y = ½ e^(tan⁻¹x) + c e^(−tan⁻¹x)

✓ Answer checked by Claude Opus 5.5 (high effort)

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