Step-by-step Solution

1. Identify and Solve the Differential Equation

The given differential equation is: $\frac{dy}{dx} + 12y = \cos\left(\frac{\pi}{12}x\right)$ This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 12$ and $Q(x) = \cos\left(\frac{\pi}{12}x\right)$.

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$. $\text{I.F.} = e^{\int 12 dx} = e^{12x}$ The general solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$. $y e^{12x} = \int e^{12x} \cos\left(\frac{\pi}{12}x\right) dx + C$ We use the standard integral formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2}(a \cos(bx) + b \sin(bx))$. Here, $a=12$ and $b=\frac{\pi}{12}$. So, $a^2 + b^2 = 12^2 + \left(\frac{\pi}{12}\right)^2 = 144 + \frac{\pi^2}{144} = \frac{144^2 + \pi^2}{144}$.

$\int e^{12x} \cos\left(\frac{\pi}{12}x\right) dx = \frac{e^{12x}}{144 + \frac{\pi^2}{144}} \left(12 \cos\left(\frac{\pi}{12}x\right) + \frac{\pi}{12} \sin\left(\frac{\pi}{12}x\right)\right)$ Substituting this back into the solution equation: $y e^{12x} = \frac{144 e^{12x}}{144^2 + \pi^2} \left(12 \cos\left(\frac{\pi}{12}x\right) + \frac{\pi}{12} \sin\left(\frac{\pi}{12}x\right)\right) + C$ Dividing by $e^{12x}$, we get the general solution for $y(x)$: $y(x) = \frac{1}{144^2 + \pi^2} \left(1728 \cos\left(\frac{\pi}{12}x\right) + 12\pi \sin\left(\frac{\pi}{12}x\right)\right) + C e^{-12x}$

2. Apply the Initial Condition

We are given the initial condition $y(0)=0$. Substituting $x=0$ and $y=0$: $0 = \frac{1}{144^2 + \pi^2} \left(1728 \cos(0) + 12\pi \sin(0)\right) + C e^{0}$ $0 = \frac{1728}{144^2 + \pi^2} + C$ $C = -\frac{1728}{144^2 + \pi^2}$ So, the particular solution is: $y(x) = \frac{1}{144^2 + \pi^2} \left(1728 \cos\left(\frac{\pi}{12}x\right) + 12\pi \sin\left(\frac{\pi}{12}x\right) - 1728 e^{-12x}\right)$

3. Analyze the Options

The solution $y(x)$ consists of two parts: a sinusoidal (oscillating) part and an exponential decay part. Let $y_{ss}(x) = \frac{1}{144^2 + \pi^2} \left(1728 \cos\left(\frac{\pi}{12}x\right) + 12\pi \sin\left(\frac{\pi}{12}x\right)\right)$ and $y_t(x) = -\frac{1728}{144^2 + \pi^2} e^{-12x}$. So, $y(x) = y_{ss}(x) + y_t(x)$.

A: $y(x)$ is an increasing function & B: $y(x)$ is a decreasing function To check for monotonicity, we examine the derivative $\frac{dy}{dx}$. From the original equation, $\frac{dy}{dx} = \cos\left(\frac{\pi}{12}x\right) - 12y$.

At $x=0$, $y(0)=0$, so $\frac{dy}{dx}\bigg|_{x=0} = \cos(0) - 12(0) = 1 > 0$. The function is increasing at $x=0$. This immediately falsifies option B.

As $x \to \infty$, the transient term $y_t(x)$ approaches 0, so $y(x)$ approaches $y_{ss}(x)$. The function $y_{ss}(x)$ is a sinusoidal function, which is not monotonic. It oscillates, so its derivative takes both positive and negative values. Since $y(x)$'s behavior for large $x$ is dominated by $y_{ss}(x)$, $y(x)$ will also oscillate and is therefore not a monotonic function. Thus, options A and B are both false.

D: $y(x)$ is a periodic function A function $f(x)$ is periodic if there exists a $T>0$ such that $f(x+T) = f(x)$ for all $x$. The term $y_{ss}(x)$ is periodic. However, the term $y_t(x) = C e^{-12x}$ (with $C \neq 0$) is not periodic. The sum of a periodic function and a non-zero non-periodic function is not periodic. Therefore, $y(x)$ is not a periodic function. Option D is false.

C: There exists a real number $\beta$ such that the line $y=\beta$ intersects the curve $y=y(x)$ at infinitely many points As $x \to \infty$, $y(x)$ approaches the steady-state solution $y_{ss}(x)$. The function $y_{ss}(x)$ is a sinusoidal function of the form $A \cos(\omega x - \phi)$. Its amplitude is: $A_{max} = \frac{\sqrt{1728^2 + (12\pi)^2}}{144^2 + \pi^2} = \frac{\sqrt{12^2(144^2 + \pi^2)}}{144^2 + \pi^2} = \frac{12\sqrt{144^2 + \pi^2}}{144^2 + \pi^2} = \frac{12}{\sqrt{144^2 + \pi^2}}$ Since $144^2+\pi^2 > 0$, the amplitude $A_{max}$ is a positive finite number. The function $y_{ss}(x)$ oscillates between $-A_{max}$ and $A_{max}$.

As $x \to \infty$, the graph of $y(x)$ becomes arbitrarily close to the graph of $y_{ss}(x)$. The local maxima of $y(x)$ converge to $A_{max}$ and the local minima converge to $-A_{max}$.

Consider any real number $\beta$ in the open interval $(-A_{max}, A_{max})$. For sufficiently large $x$, the local maxima of $y(x)$ will be greater than $\beta$, and the local minima will be less than $\beta$. By the Intermediate Value Theorem, between each successive local maximum and minimum, the continuous function $y(x)$ must take the value $\beta$. Since there are infinitely many local maxima and minima as $x \to \infty$, the line $y=\beta$ will intersect the curve $y=y(x)$ at infinitely many points. Therefore, statement C is true.

Conclusion

Based on the analysis:

• Statement A is FALSE.
• Statement B is FALSE.
• Statement C is TRUE.
• Statement D is FALSE.

The only true statement is C.