Question

Solve: \[ (1+x^2) \frac{dy}{dx} + y = \tan^{-1} x. \]

Hint:

Use the integrating factor method for first-order linear differential equations and substitution for tricky integrals.

Answer 1

Rewrite the equation: \[ (1+x^2) \frac{dy}{dx} + y = \tan^{-1} x \implies \frac{dy}{dx} + \frac{y}{1+x^2} = \frac{\tan^{-1} x}{1+x^2}. \] This is a linear first-order differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x), \] where \[ P(x) = \frac{1}{1+x^2}, \quad Q(x) = \frac{\tan^{-1} x}{1+x^2}. \] The integrating factor (IF) is: \[ \mu(x) = e^{\int P(x) dx} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1} x}. \] Multiply both sides by the integrating factor: \[ e^{\tan^{-1} x} \frac{dy}{dx} + \frac{e^{\tan^{-1} x}}{1+x^2} y = \frac{\tan^{-1} x}{1+x^2} e^{\tan^{-1} x}. \] Left side is: \[ \frac{d}{dx} \left( y \, e^{\tan^{-1} x} \right) = \frac{\tan^{-1} x}{1+x^2} e^{\tan^{-1} x}. \] Integrate both sides: \[ y \, e^{\tan^{-1} x} = \int \frac{\tan^{-1} x}{1+x^2} e^{\tan^{-1} x} \, dx + C. \] Let \[ t = \tan^{-1} x \implies dt = \frac{1}{1+x^2} dx. \] So, \[ \int \frac{\tan^{-1} x}{1+x^2} e^{\tan^{-1} x} dx = \int t e^{t} dt. \] Integrate by parts: \[ \int t e^{t} dt = t e^{t} - \int e^{t} dt = t e^{t} - e^{t} + C = e^{t} (t - 1) + C. \] Therefore, \[ y \, e^{\tan^{-1} x} = e^{\tan^{-1} x} (\tan^{-1} x - 1) + C, \] and \[ \boxed{ y = \tan^{-1} x - 1 + C e^{-\tan^{-1} x}. } \]