(a) The differential equation
[tex]y' + \dfrac14 y = 3 + 2 \cos(2x)[/tex]
is linear, so we can use the integrating factor method. We have I.F.
[tex]\mu = \displaystyle \exp\left(\int \frac{dx}4\right) = e^{x/4}[/tex]
so that multiplying both sides by [tex]\mu[/tex] gives
[tex]e^{x/4} y' + \dfrac14 e^{x/4} y = 3e^{x/4} + 2 e^{x/4} \cos(2x)[/tex]
[tex]\left(e^{x/4} y\right)' = 3e^{x/4} + 2 e^{x/4} \cos(2x)[/tex]
Integrate both sides. (Integrate by parts twice on the right side; I'll omit the details.)
[tex]e^{x/4} y = 12 e^{x/4} + \dfrac8{65} e^{x/4} (8\sin(2x) + \cos(2x)) + C[/tex]
Solve for [tex]y[/tex].
[tex]y = 12 + \dfrac8{65} (\sin(2x) + \cos(2x)) + Ce^{-x/4}[/tex]
Given that [tex]y(0)=0[/tex], we find
[tex]0 = 12 + \dfrac8{65} (\sin(0) + \cos(0)) + Ce^0 \implies C = -\dfrac{788}{65}[/tex]
and the particular solution to the initial value problem is
[tex]\boxed{y = 12 + \dfrac8{65} (\sin(2x) + \cos(2x)) - \dfrac{788}{65} e^{-x/4}}[/tex]
As [tex]x[/tex] gets large, the exponential term will converge to 0. We have
[tex]\sin(2x) + \cos(2x) = \sqrt2 \sin\left(2x + \dfrac\pi4\right)[/tex]
which means the trigonometric terms will oscillate between [tex]\pm\sqrt2[/tex]. So overall, the solution will oscillate between [tex]12\pm\sqrt2[/tex] for large [tex]x[/tex].
(b) We want the smallest [tex]x[/tex] such that [tex]y=12[/tex], i.e.
[tex]0 = \dfrac8{65} (\sin(2x) + \cos(2x)) - \dfrac{788}{65} e^{-x/4}[/tex]
[tex]\dfrac{788}{65} e^{-x/4} = \dfrac{8\sqrt2}{65} \sin\left(2x + \dfrac\pi4\right)[/tex]
[tex]\dfrac{197}{\sqrt2} e^{-x/4} = \sin\left(2x + \dfrac\pi4\right)[/tex]
Using a calculator, the smallest solution seems to be around [tex]\boxed{x\approx21.909}[/tex]
