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Analysis of Solution to First Order Linear Differential Equation

For x∈Rx \in \mathbb{R}, let the function y(x)y(x) be the solution of the differential equation

dydx+12y=cos⁡(π12x),y(0)=0.\frac{dy}{dx} + 12y = \cos\left(\frac{\pi}{12} x\right), \quad y(0) = 0.

Then, which of the following statements is/are TRUE ?

Options

A

y(x)y(x) is an increasing function

B

y(x)y(x) is a decreasing function

C

There exists a real number β\beta such that the line y=βy = \beta intersects the curve y=y(x)y = y(x) at infinitely many points

Correct
D

y(x)y(x) is a periodic function

Step-by-Step Solution

To analyze the given first-order linear differential equation dydx+12y=cos⁡(π12x),y(0)=0\frac{dy}{dx} + 12y = \cos\left(\frac{\pi}{12} x\right), \quad y(0) = 0

we compute the integrating factor I(x)=e∫12 dx=e12xI(x) = e^{\int 12 \, dx} = e^{12x}. Solving the differential equation yields the explicit solution: y(x)=1144+π2144(12cos⁡(π12x)+π12sin⁡(π12x)−12e−12x)y(x) = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos\left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin\left(\frac{\pi}{12} x\right) - 12 e^{-12x}\right)

Analyzing the properties of y(x)y(x):

  1. Monotonicity (Options A and B): The solution contains sinusoidal terms that cause y(x)y(x) to oscillate infinitely as x→∞x \to \infty, making y(x)y(x) neither strictly increasing nor strictly decreasing on R\mathbb{R}.
  2. Periodicity (Option D): The presence of the transient decaying exponential term −12e−12x-12 e^{-12x} breaks the periodicity of the function, so y(x)y(x) is not periodic.
  3. Intersections with y=βy = \beta (Option C): For β=0\beta = 0, the equation y(x)=0y(x) = 0 simplifies to an equation where the oscillating sinusoidal component balances the exponentially decaying term, yielding infinitely many real solutions as x→∞x \to \infty.

Thus, statement (C) is the correct choice.

Correct Answer: (C)

Analysis of Solution to First Order Linear Differential Equation | Mathematics PYQ Solution - JEE Challenger