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Class 12 Differential Equations · Very Long Answer5 Markslinear_differential_equations

Class 12 Maths · Differential Equations · Very Long Answer #109

Solve (1 + x²)y′ = e^(m arctan x) − y, y(0) = 1, for a real constant m.

✅ Step-by-Step Solution

Final Answer: For m ≠ −1: ye^(arctan x) = [e^((m+1)arctan x) + m]/(m + 1). For m = −1: y = (1 + arctan x)e^(−arctan x).
  1. Normalize to y′ + y/(1 + x²) = e^(m arctan x)/(1 + x²). Use IF e^(arctan x).
  2. Set t = arctan x. Then d(yeᵗ)/dt = e^((m + 1)t).
  3. If m ≠ −1, integrate yeᵗ = e^((m + 1)t)/(m + 1) + C. The condition at t = 0 gives C = m/(m + 1).
  4. If m = −1, integrate yeᵗ = t + C; the initial condition gives C = 1.
  5. Thus for m ≠ −1, ye^(arctan x) = [e^((m+1)arctan x) + m]/(m + 1); for m = −1, y = (1 + arctan x)e^(−arctan x), x ∈ ℝ.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Normalize to y′ + y/(1 + x²) = e^(m arctan x)/(1 + x²). Use IF e^(arctan x).
  • [1 mark] Set t = arctan x. Then d(yeᵗ)/dt = e^((m + 1)t).
  • [1 mark] If m ≠ −1, integrate yeᵗ = e^((m + 1)t)/(m + 1) + C. The condition at t = 0 gives C = m/(m + 1).
  • [1 mark] If m = −1, integrate yeᵗ = t + C; the initial condition gives C = 1.
  • [1 mark] Thus for m ≠ −1, ye^(arctan x) = [e^((m+1)arctan x) + m]/(m + 1); for m = −1, y = (1 + arctan x)e^(−arctan x), x ∈ ℝ.

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