NCERT Class 12 Maths · Chapter 9 Differential Equations · Exercise 9.5 Question 8

✅ Step-by-Step Solution

1. Write y′ + \(\frac{2 x}{x^{2} + 1}\) y = \(\frac{\cot{\left(x \right)}}{x^{2} + 1}\); work away from sin x=0.
2. An integrating factor is μ=\(x^{2} + 1\), since μ′/μ=\(\frac{2 x}{x^{2} + 1}\).
3. Multiply through: d(μy)/dx = \(\cot{\left(x \right)}\).
4. Integrate the right side: \(\log{\left(\left|{\sin{\left(x \right)}}\right| \right)}\). Thus μy=\(\log{\left(\left|{\sin{\left(x \right)}}\right| \right)}\)+C.
5. Therefore y = \(\frac{C + \log{\left(\left|{\sin{\left(x \right)}}\right| \right)}}{x^{2} + 1}\).