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

$\dfrac{dy}{dx} + 12y = \cos\!\left(\dfrac{\pi}{12}x\right), \qquad y(0) = 0$ $\text{To find: which of the four statements are TRUE}$ $1.\ \text{solve the linear DE (integrating factor)}$ $2.\ \text{write the explicit } y(x)$ $3.\ \text{test the four statements}$ $\dfrac{dy}{dx} + 12y = \cos\!\left(\dfrac{\pi}{12}x\right) \Rightarrow \text{IF} = e^{\int 12\,dx} = e^{12x}$ $\dfrac{d}{dx}\left(y\,e^{12x}\right) = e^{12x}\cos\!\left(\dfrac{\pi}{12}x\right)$ $y e^{12x} = \displaystyle\int e^{12x}\cos\!\left(\dfrac{\pi}{12}x\right)dx = e^{12x}\cdot\dfrac{12\cos\!\left(\frac{\pi x}{12}\right) + \frac{\pi}{12}\sin\!\left(\frac{\pi x}{12}\right)}{D} + C$ $\text{where } D = 144 + \dfrac{\pi^{2}}{144}$ $y(x) = \dfrac{12\cos\!\left(\frac{\pi x}{12}\right) + \frac{\pi}{12}\sin\!\left(\frac{\pi x}{12}\right)}{D} + C e^{-12x}$ $y(0) = \dfrac{12}{D} + C = 0 \Rightarrow C = -\dfrac{12}{D}$ $y(x) = y_p(x) - \dfrac{12}{D}e^{-12x}, \quad y_p(x) = \dfrac{12\cos\!\left(\frac{\pi x}{12}\right) + \frac{\pi}{12}\sin\!\left(\frac{\pi x}{12}\right)}{D}$ $x \to \infty:\ e^{-12x} \to 0 \Rightarrow y \to y_p \ (\text{period } 24,\ \text{amplitude } \tfrac{1}{\sqrt D})$ $\Rightarrow \text{for } |\beta| < \tfrac{1}{\sqrt D},\ y = \beta \text{ meets } y(x) \text{ infinitely often} \Rightarrow (3)$ $y'(x) = \cos\!\left(\frac{\pi x}{12}\right) - 12y(x):\quad y'(0) = 1 > 0, \ y'(12) < 0$ $\Rightarrow y \text{ is neither increasing nor decreasing} \Rightarrow (1), (2) \text{ FALSE}$ $y = y_p - \dfrac{12}{D}e^{-12x};\ \text{the transient never vanishes} \Rightarrow \text{not periodic} \Rightarrow (4) \text{ FALSE}$ $\boxed{\text{Answer: only (3)}}$