Question

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

• A. $y(x)$ is an increasing function
• B. $y(x)$ is a decreasing function
• C. There exists a real number $\beta$ such that the line $y=\beta$ intersects the curve $y=y(x)$ at infinitely many points
• D. $y(x)$ is a periodic function

Answer 1

Answer: (C)

$\frac{dy}{dx}+12y=\cos \left(\frac{\pi }{12}x\right)$
Linear D.E.
$\text{ I.F. }=e^{\int 12.dx}=e^{12x}$
Solution of DE
$y\cdot e^{12x}=\int e^{12x}\cdot \cos \left(\frac{\pi }{12}x\right)dx$
$y\cdot e^{12x}=\frac{e^{12x}}{(12{)}^2+{\left(\frac{\pi }{12}\right)}^2}\left(12\cos \frac{\pi }{12}x+\frac{\pi }{12}\sin \frac{\pi }{12}x\right)+C$
$\Rightarrow y=\frac{(12)}{(12{)}^4+{\pi }^2}\left((12{)}^2\cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)\right)+\frac{C}{e^{12x}}$
Given $y(0)=0$
$\Rightarrow 0=\frac{12}{12^4+{\pi }^2}\left(12^2+0\right)+C\Rightarrow C=\frac{-12^3}{12^4+{\pi }^2}$
$\therefore y=\frac{12}{12^4+{\pi }^2}\left[(12{)}^2\cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)-12^2\cdot e^{-12x}\right]$
Now $\frac{dy}{dx}=\frac{12}{12^4+{\pi }^2}[\underset{\text{min. value }}{\underbrace{-12\pi \sin \left(\frac{\pi x}{12}\right)+\frac{{\pi }^2}{12}\cos \left(\frac{\pi x}{12}\right)}}+12^3{e}^{-12x}]$
$(-\sqrt{144{\pi }^2+\frac{{\pi }^4}{144}}=-12\pi \sqrt{1+\frac{{\pi }^2}{12^4})}$
$\Rightarrow \frac{dy}{dx}>0\forall x\le 0\text{ \& may be negative/positive for }x>0$
So, $f(x)$ is neither increasing nor decreasing
For some $\beta \in R,y=\beta$ intersects $y=f(x)$ at infinitely many points
So option $C$ is correct