Class 12 Differential Equations · Long Answer3 Markslinear_differential_equations
Class 12 Maths · Differential Equations · Long Answer #61
Solve x ln x · y′ + y = (2/x)ln x.
✅ Step-by-Step Solution
Final Answer: y ln x = −2(1 + ln x)/x + C for x > 0, x ≠ 1.
- For x > 0, x ≠ 1, divide by x ln x: y′ + y/(x ln x) = 2/x². Use IF ln x.
- Then (y ln x)′ = 2ln x/x².
- By parts, ∫2ln x/x²dx = −2ln x/x − 2/x + C.
- Thus y ln x = −2(1 + ln x)/x + C. The smooth extension across x = 1 exists only for C = 2 and has y(1) = 0; the original equation at x = 1 requires this value.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] For x > 0, x ≠ 1, divide by x ln x: y′ + y/(x ln x) = 2/x². Use IF ln x.
- [1 mark] Then (y ln x)′ = 2ln x/x².
- [1 mark] By parts, ∫2ln x/x²dx = −2ln x/x − 2/x + C.
- [0.5 marks] Thus y ln x = −2(1 + ln x)/x + C. The smooth extension across x = 1 exists only for C = 2 and has y(1) = 0; the original equation at x = 1 requires this value.
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