Question

For $x \in R$, let the function $y ( x )$ be the solution of the differential equation
$\frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right), y(0)=0 \text {. }$
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)

$\frac{ dy }{ dx }+12 y =\cos \left(\frac{\pi}{12} x\right)$
Linear D.E.
$\text { I.F. }= e ^{\int 12 . d x}= e ^{12 x}$
Solution of DE
$y \cdot e^{12 x}=\int e^{12 x} \cdot \cos \left(\frac{\pi}{12} x\right) d x $
$ y \cdot e^{12 x}=\frac{e^{12 x}}{(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$
$\Rightarrow y=\frac{(12)}{(12)^4+\pi^2}\left((12)^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)\right)+\frac{C}{e^{12 x}}$
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 ^{-12 x }\right] $
Now $ \frac{ dy }{ dx }=\frac{12}{12^4+\pi^2}[\underbrace{-12 \pi \sin \left(\frac{\pi x }{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x }{12}\right)}_{\text {min. value }}+12^3 e ^{-12 x }] $
$ \left(-\sqrt{144 \pi^2+\frac{\pi^4}{144}}=-12 \pi \sqrt{\left.1+\frac{\pi^2}{12^4}\right)}\right.$
$ \Rightarrow \frac{ dy }{ dx }>0 \forall x \leq 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