Answer: The general solution is \(x=Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t\).

As \(t\to\infty\), the terms \(e^{-2t}\) and \(te^{-2t}\) both tend to \(0\), so \(x\) approaches \(-\frac{1}{8}\cos 2t\). Hence \(x\) oscillates with constant amplitude \(\frac{1}{8}\) about \(0\).

We first solve the homogeneous equation

\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+4\frac{\mathrm{d}x}{\mathrm{d}t}+4x=0\).

Its auxiliary equation is \(m^2+4m+4=0\), so \((m+2)^2=0\). Hence the complementary function is

\(x_c=Ae^{-2t}+Bte^{-2t}\).

For a particular solution, since the forcing term is \(\sin 2t\), take

\(x_p=p\sin 2t+q\cos 2t\).

Then

\(\frac{\mathrm{d}x_p}{\mathrm{d}t}=2p\cos 2t-2q\sin 2t\),

\(\frac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=-4p\sin 2t-4q\cos 2t\).

Substitute into the differential equation:

\((-4p\sin 2t-4q\cos 2t)+4(2p\cos 2t-2q\sin 2t)+4(p\sin 2t+q\cos 2t)=\sin 2t.\)

Collecting terms gives

\(-8q\sin 2t+8p\cos 2t=\sin 2t.\)

So

\(-8q=1\) and \(8p=0\),

hence \(q=-\frac{1}{8}\) and \(p=0\). Therefore

\(x_p=-\frac{1}{8}\cos 2t\).

So the general solution is

\(x=Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t\).

As \(t\to\infty\), both \(e^{-2t}\to 0\) and \(te^{-2t}\to 0\), so the transient terms decay away. Thus

\(x\sim -\frac{1}{8}\cos 2t\) as \(t\to\infty\).

Therefore \(x\) oscillates indefinitely between \(\pm \frac{1}{8}\), with the exponential terms disappearing over time.