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

Given,

$\frac{dy}{dx}+12y=\cos \left(\frac{\pi x}{12}\right)$

$I.F=e^{12x}$

Now solution of the differential equation is given by,

$\Rightarrow y·e^{12x}=\int e^{12x}·\cos \left(\frac{\pi x}{12}\right)dx+C$

$\Rightarrow y·e^{12x}=\frac{e^{12x}}{{12}^2+{\left(\frac{\pi }{12}\right)}^2}\left[12\cos \frac{\pi x}{12}+\frac{\pi }{12}\sin \left(\frac{\pi x}{12}\right)\right]+C$

Now given curve is passing through $\left(0,0\right)$

So, $y\left(0\right)=0\Rightarrow C=-\frac{12}{{12}^2+{\left(\frac{\pi }{12}\right)}^2}$

So $y=\frac{1}{\lambda }\left[\underset{f_1\left(x\right)}{\underbrace{12\cos \left(\frac{\pi x}{12}\right)+\frac{\pi }{12}\sin \left(\frac{\pi x}{12}\right)}}-12e^{-12x}\right]$

{Where $\lambda =\frac{1}{{12}^2+{\left(\frac{\pi }{12}\right)}^2}$ }

Now on differentiating we get,

$\frac{dy}{dx}=\frac{1}{{\lambda }^{\text{'}}}\left[\underset{f_2\left(x\right)}{\underbrace{-\pi \sin \left(\frac{\pi x}{12}\right)+\frac{{\pi }^2}{{12}^2}\cos \left(\frac{\pi x}{12}\right)}}+12e^{-12x}\right]$

$\left\{\text{where} {\lambda }^{\text{'}}=\frac{12}{{12}^2+{\left(\frac{\pi }{12}\right)}^2}\right\}$

Now when $x$ is large then $12e^{-12x}$ tends to zero.

But $f_2\left(x\right)$ varies in $\left[-\sqrt{{\pi }^2+{\left(\frac{\pi }{12}\right)}^4} , \sqrt{{\pi }^2+{\left(\frac{\pi }{12}\right)}^4}\right]$ $\left\{\text{by using }-\sqrt{a^2+b^2} \& \sqrt{a^2+b^2}\right\}$

Hence $\frac{dy}{dx}$ is changing its sign.

So $y\left(x\right)$ is non monotonic for all real number.
Also when $x$ is very large then again $-12e^{-12x}$ is almost zero but $f_1\left(x\right)$ is periodic, so there exist some $\beta$ for which $y=\beta$ intersect $y=y\left(x\right)$ at infinitely many points.