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NDA 2017 Maths · Mathematics · Differential Equations · medium

Eliminating the arbitrary constants A and B from y=Acos 2x+Bsin 2x gives the differential equation

Single correct option

Eliminating the arbitrary constants AA and BB from y=Acos⁡2x+Bsin⁡2xy=A\cos2x+B\sin2x gives the differential equation
  1. A. y′′+4y=0y''+4y=0
  2. B. y′′−4y=0y''-4y=0
  3. C. y′′+2y=0y''+2y=0
  4. D. y′+4y=0y'+4y=0
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Answer: (A) y′′+4y=0y''+4y=0

Differentiating twice, y′′=−4Acos⁡2x−4Bsin⁡2x=−4yy''=-4A\cos2x-4B\sin2x=-4y, so y′′+4y=0y''+4y=0. Two constants mean order 22, which rules out y′+4y=0y'+4y=0. The factor is 22=42^2=4, not 22, and trigonometric solutions need the plus sign (the minus sign belongs to exponentials).
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Original question written in the NDA 2017 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake