Answer :
In the differential equation
[tex]\dfrac{dy}{dx} + 12y = \cos\left(\dfrac{\pi x}{12}\right)[/tex]
multiply on both sides by the integrating factor
[tex]\mu = \exp\left(\displaystyle\int12\,dx\right) = e^{12x}[/tex]
Then the left side condenses to the derivative of a product.
[tex]e^{12x} \dfrac{dy}{dx} + 12 e^{12x} y = e^{12x} \cos\left(\dfrac{\pi x}{12}\right)[/tex]
[tex]\dfrac{d}{dx}\left[e^{12x}y\right] = e^{12x}\cos\left(\dfrac{\pi x}{12}\right)[/tex]
Integrate both sides with respect to [tex]x[/tex], and use the initial condition [tex]y(0)=0[/tex] to solve for the constant [tex]C[/tex].
[tex]\displaystyle \int \frac{d}{dx} \left[e^{12x}y\right] \, dx = \int e^{12x} \cos\left(\dfrac{\pi x}{12}\right) \, dx[/tex]
As an alternative to integration by parts, recall
[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]
Now
[tex]e^{12x} \cos\left(\dfrac{\pi x}{12}\right) = e^{12x} \mathrm{Re}\left(e^{i\pi x/12}\right) = \mathrm{Re}\left(e^{(12+i\pi/12)x}\right)[/tex]
[tex]\displaystyle \int \mathrm{Re}\left(e^{(12+i\pi/12)x}\right) \, dx = \mathrm{Re}\left(\int e^{(12+i\pi/12)x} \, dx\right)[/tex]
[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac1{12+i\frac\pi{12}} e^{(12+i\pi/12)x}\right) + C[/tex]
[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac{12 - i\frac\pi{12}}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \left(\cos\left(\frac{\pi x}{12}\right) + i \sin\left(\frac{\pi x}{12}\right)\right)\right) + C[/tex]
[tex]\displaystyle. ~~~~~~~~ = \frac{12}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \cos\left(\frac{\pi x}{12}\right) + \frac\pi{12} e^{12x} \sin\left(\frac{\pi x}{12}\right) + C[/tex]
[tex]\displaystyle. ~~~~~~~~ = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]
Solve for [tex]y[/tex].
[tex]\displaystyle e^{12x} y = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]
[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]
Solve for [tex]C[/tex].
[tex]y(0)=0 \implies 0 = \dfrac1{12(12^4+\pi^2)} \left(12^4 + 0\right) + C \implies C = -\dfrac{12^3}{12^4+\pi^2}[/tex]
So, the particular solution to the initial value problem is
[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) - \frac{12^3}{12^4+\pi^2}[/tex]
Recall that
[tex]R\cos(\alpha-\beta) = R\cos(\alpha)\cos(\beta) + R\sin(\alpha)\sin(\beta)[/tex]
Let [tex]\alpha=\frac{\pi x}{12}[/tex]. Then
[tex]\begin{cases} R\cos(\beta) = 12^4 \\ R\sin(\beta) = 12^4\pi+\pi^3 \end{cases} \\\\ \implies \begin{cases} (R\cos(\beta))^2 + (R\sin(\beta))^2 = R^2 = 12^8 + (12^4\pi + \pi^3)^2 \\ \frac{R\sin(\beta)}{R\cos(\beta)}=\tan(\beta)=\pi+\frac{\pi^3}{12^4}\end{cases}[/tex]
Whatever [tex]R[/tex] and [tex]\beta[/tex] may actually be, the point here is that we can condense [tex]y[/tex] into a single cosine expression, so choice (D) is correct, since [tex]\cos(x)[/tex] is periodic. This also means choice (C) is also correct, since [tex]\beta=\cos(x)\implies\beta=\cos(x+2n\pi)[/tex] for infinitely many integers [tex]n[/tex]. This simultaneously eliminates (A) and (B).
