JEE Advanced 2022 Paper 2 Q14 Mathematics Integration & Differential Equations Linear Differential Equations Hard

JEE Advanced 2022 Paper 2 · Q14 · Linear Differential Equations

For $x\in\mathbb{R}$, let the function $y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}+12y = \cos\left(\dfrac{\pi}{12}x\right)$, $y(0)=0$. Then, which of the following statements is/are TRUE?

  1. A. $y(x)$ is an increasing function
  2. B. $y(x)$ is a decreasing function
  3. C. There exists a real number $\beta$ such that the line $y=\beta$ intersects the curve $y=y(x)$ at infinitely many points
  4. D. $y(x)$ is a periodic function
Reveal answer + step-by-step solution

Correct answer:C

Solution

Linear ODE with integrating factor $e^{12x}$. Solution: $y(x)e^{12x} = \int e^{12x}\cos(\pi x/12)\,dx + C$. The particular integral is of the form $A\cos(\pi x/12)+B\sin(\pi x/12)$ (bounded periodic). Specifically: $y(x) = \dfrac{1}{144+\pi^2/144}\left(12\cos(\pi x/12)+(\pi/12)\sin(\pi x/12)\right) + C e^{-12x}$, with $C$ chosen so $y(0)=0$. As $x\to\infty$, the transient $Ce^{-12x}\to 0$ and $y(x)$ approaches a bounded periodic function with period $24$. So $y(x)$ is neither monotonic increasing nor decreasing (A and B false), and not exactly periodic for all $x$ because of the transient (D false). However, the bounded oscillation means there is a horizontal line $y=\beta$ (any value strictly between the asymptotic min and max) that crosses the curve infinitely many times, so (C) true.

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