GATE AG 2021 • Q36
MCQ • 2 Marks
Section 1: Engineering Mathematics
Differential Equations
Subtopic: Higher order linear ODEs with constant coefficients
Question Statement:
Solution of the differential equation y″ + y′ + 0.25y = 0 with the initial values y(0) = 3.0 and y′(0) = −3.5 is
Options:
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(A) y = (3 − 2x)e^0.5x
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(B) y = (3 − 2x)e^(−0.25x)
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(C) y = (3 − 2x)e^(−0.5x)
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(D) y = (2 − 3x)e^(−0.5x)
Step-by-Step Solution & Explanation
(A) y = (3 − 2x)e^0.5x
(B) y = (3 − 2x)e^(−0.25x)
(C) y = (3 − 2x)e^(−0.5x)
(D) y = (2 − 3x)e^(−0.5x)
Characteristic equation: r² + r + 0.25 = 0 → (r+0.5)² = 0, a repeated root r = −0.5. General solution: y = (C₁ + C₂x)e^(−0.5x). Applying y(0)=3: C₁ = 3. Differentiating: y′ = C₂e^(−0.5x) − 0.5(C₁+C₂x)e^(−0.5x); at x=0: y′(0) = C₂ − 0.5C₁ = −3.5 → C₂ = −3.5+1.5 = −2. So y = (3 − 2x)e^(−0.5x), matching (C).
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