Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
If $y = e^{-x}\sin x$ and $y_n + a_ny_{n-1} = 0$ where an is constant for $n \in \mathbb{N}$ & $y_n = \frac{d^ny}{dx^n}$ (nth derivative of $y$), then:
a_4 = 4
a_8 = -16
a_12 = 64
a_16 = 256
Step-by-Step Solution
Key Concept: Successive differentiation of exponential-trigonometric functions reveals periodic patterns in derivatives that lead to higher-order differential equations.
Starting with $y = e^{-x}\sin x$, successive derivatives give $y_2 = e^{-x}\cos x - e^{-x}\sin x = (\sqrt{2})e^{-x}\sin(x - \frac{\pi}{2})$ and $y_1 = \sqrt{2}e^{-x}\sin(\frac{\pi}{4} - x)$. Computing further derivatives: $y_3 = -(\sqrt{2})^{2/3}e^{-x}\sin(x - \frac{3\pi}{4})$ and $y_4 = -4e^{-x}\sin(x - \pi) = -4y$. Therefore $y_4 + 4y = 0$ implies $y_3 + 4y_3 = 0$, giving the differential equation $y_3 - 16y_1 = 0$.
Correct Answer: 1,2,3