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Class 12 Differential Equations · Very Long Answer5 Markshomogeneous_differential_equations

Class 12 Maths · Differential Equations · Very Long Answer #113

Solve x²dy + (xy + y²)dx = 0, y(1) = 1.

✅ Step-by-Step Solution

Final Answer: y = 2x/(3x² − 1), x > 1/√3.
  1. For x ≠ 0 put v = y/x. Then xv′ = −2v − v².
  2. Near v = 1, separate dv/[v(v + 2)] = −dx/x.
  3. Using 1/[v(v + 2)] = ½(1/v − 1/(v + 2)), integrate ln|v/(v + 2)| = −2ln|x| + C.
  4. At (1, 1), v/(v + 2) = 1/3, giving v/(v + 2) = 1/(3x²). Thus y = 2x/(3x² − 1).
  5. Therefore y = 2x/(3x² − 1), on (1/√3, ∞), the maximal connected interval containing 1.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] For x ≠ 0 put v = y/x. Then xv′ = −2v − v².
  • [1 mark] Near v = 1, separate dv/[v(v + 2)] = −dx/x.
  • [1 mark] Using 1/[v(v + 2)] = ½(1/v − 1/(v + 2)), integrate ln|v/(v + 2)| = −2ln|x| + C.
  • [1 mark] At (1, 1), v/(v + 2) = 1/3, giving v/(v + 2) = 1/(3x²). Thus y = 2x/(3x² − 1).
  • [1 mark] Therefore y = 2x/(3x² − 1), on (1/√3, ∞), the maximal connected interval containing 1.

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