Class 12 Differential Equations · Very Long Answer5 Markslinear_differential_equations
Class 12 Maths · Differential Equations · Very Long Answer #77
Solve xy′ + y − x + xy cot x = 0, x ≠ 0.
✅ Step-by-Step Solution
Final Answer: y = 1/x − cot x + C/(x sin x), x sin x ≠ 0.
- Divide by x: y′ + (1/x + cot x)y = 1, requiring x ≠ 0 and sin x ≠ 0.
- A convenient integrating factor is x sin x, nonzero on an admissible connected interval.
- Then (xy sin x)′ = x sin x.
- By parts, ∫x sin x dx = −x cos x + sin x + C.
- Therefore y = 1/x − cot x + C/(x sin x), with x sin x ≠ 0.
📋 CBSE Stepwise Marking Scheme
- [1 mark] Divide by x: y′ + (1/x + cot x)y = 1, requiring x ≠ 0 and sin x ≠ 0.
- [1 mark] A convenient integrating factor is x sin x, nonzero on an admissible connected interval.
- [1 mark] Then (xy sin x)′ = x sin x.
- [1 mark] By parts, ∫x sin x dx = −x cos x + sin x + C.
- [1 mark] Therefore y = 1/x − cot x + C/(x sin x), with x sin x ≠ 0.
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