Question

Find the particular solution of: \[ (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}, y = 0 \text{ when } x = 1. \]

Hint:

Always write differential equation in linear form, find I.F, then integrate.

Answer 1

Step 1: Write in standard linear form: \[ \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2}. \] Step 2: Integrating Factor (I.F): \[ I.F = \exp\left( \int \frac{2x}{1 + x^2} dx \right) = \exp[ \ln(1 + x^2) ] = 1 + x^2. \] Step 3: General solution: \[ y \cdot I.F = \int RHS \cdot I.F\, dx + C. \] So, \[ y (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) dx + C = \int \frac{1}{1 + x^2} dx + C = \tan^{-1} x + C. \] So, \[ y = \frac{ \tan^{-1} x + C }{1 + x^2}. \] Step 4: Apply $y = 0$ when $x = 1$: \[ 0 = \frac{ \tan^{-1} 1 + C }{2} \implies 0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}. \] Final answer: \[ \boxed{ y = \frac{ \tan^{-1} x - \frac{\pi}{4} }{ 1 + x^2 } }. \] %Quciktip