<p>If \(y = e^{-x}\cos x\) and \(y_n + k_n y = 0\), where \(y_n = \dfrac{d^n y}{dx^n}\) and \(k, n \in N\) are constants</p>
Step-by-Step Solution
Key Concept: Find the recurrence relation for successive derivatives of y = e^(-x)cos(x) by recognizing the pattern in derivatives, then identify the constant k that makes y_n + ky = 0 for a specific n.
<p><strong>Step 1:</strong> Find the first few derivatives of y = e^(-x)cos(x)</p><p>y = e^(-x)cos(x)</p><p>y₁ = e^(-x)(-cos x - sin x) = -e^(-x)(cos x + sin x)</p><p>y₂ = e^(-x)(cos x + sin x) - e^(-x)(cos x + sin x) + e^(-x)(-sin x + cos x) = -2e^(-x)sin(x)</p><p>y₃ = -2[-e^(-x)sin(x) + e^(-x)cos(x)] = 2e^(-x)(sin x - cos x)</p><p>y₄ = 2[e^(-x)(sin x - cos x) + e^(-x)(cos x + sin x)] = 4e^(-x)cos(x) = 4y</p></p><p><strong>Step 2:</strong> Recognize the pattern: y₄ = 4e^(-x)cos(x) = 4y</p><p>This means: y₄ - 4y = 0</p><p><strong>Step 3:</strong> Compare with the given form y_n + ky = 0</p><p>We have y₄ + (-4)y = 0, so n = 4 and k = -4</p><p><strong>Step 4:</strong> Verify using characteristic equation method: For y = e^(-x)cos(x), the characteristic roots are (-1 ± i), giving (r+1)² + 1 = 0, or r² + 2r + 2 = 0. The 4th derivative relation is (D² + 2D + 2)² = D⁴ + 4D³ + 6D² + 4D + 1 applied to exponential-trigonometric gives: y₄ + 4y₃ + 6y₂ + 4y₁ + y = 0, but directly: y₄ = 4y</p><p>∴ Answer: k = -4 (and n = 4)</p>
Correct Answer: A