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Class 12 Differential Equations · Long Answer3 Markshomogeneous_differential_equations

Class 12 Maths · Differential Equations · Long Answer #99

Solve x²y′ − xy = 1 + cos(y/x), x ≠ 0, y(1) = π/2.

✅ Step-by-Step Solution

Final Answer: tan(y/(2x)) = 3/2 − 1/(2x²); y = 2x arctan(3/2 − 1/(2x²)), x > 0.
  1. Put v = y/x. Since x²y′ − xy = x³v′, dv/(1 + cos v) = dx/x³ near v = π/2.
  2. Use 1 + cos v = 2cos²(v/2), giving ∫dv/(1 + cos v) = tan(v/2). Thus tan(v/2) = −1/(2x²) + C.
  3. At (1, π/2), tan(π/4) = 1, so C = 3/2.
  4. Thus y = 2x arctan(3/2 − 1/(2x²)), x > 0, with the branch through the initial point.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Put v = y/x. Since x²y′ − xy = x³v′, dv/(1 + cos v) = dx/x³ near v = π/2.
  • [1 mark] Use 1 + cos v = 2cos²(v/2), giving ∫dv/(1 + cos v) = tan(v/2). Thus tan(v/2) = −1/(2x²) + C.
  • [1 mark] At (1, π/2), tan(π/4) = 1, so C = 3/2.
  • [0.5 marks] Thus y = 2x arctan(3/2 − 1/(2x²)), x > 0, with the branch through the initial point.

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