NCERT Class 12 Maths · Chapter 9 Differential Equations · Exercise 9.5 Question 15

✅ Step-by-Step Solution

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