Answer: The general solution is \(x=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)+2\sin t-\cos t\).

Using \(x(0)=5\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}(0)=2\), the particular solution is \(x=\mathrm{e}^{-t}(6\cos 2t+3\sin 2t)+2\sin t-\cos t\).

For large positive \(t\), \(\mathrm{e}^{-t}\to 0\), so \(x\approx 2\sin t-\cos t\).

First solve the associated homogeneous equation

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+2\dfrac{\mathrm{d}x}{\mathrm{d}t}+5x=0\).

Try \(x=\mathrm{e}^{mt}\). Then

\(m^2+2m+5=0\).

So

\(m=\dfrac{-2\pm \sqrt{4-20}}{2}=-1\pm 2\mathrm{i}\).

Hence the complementary function is

\(x_c=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)\).

For a particular solution, since the forcing term is \(10\sin t\), take

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

Then

\(\dot{x}_p=-p\sin t+q\cos t\),

\(\ddot{x}_p=-p\cos t-q\sin t\).

Substitute into the differential equation:

\((-p\cos t-q\sin t)+2(-p\sin t+q\cos t)+5(p\cos t+q\sin t)=10\sin t\).

Collect coefficients of \(\cos t\) and \(\sin t\):

\((4p+2q)\cos t+(-2p+4q)\sin t=0\cdot \cos t+10\sin t\).

So

\(4p+2q=0\), \(-2p+4q=10\).

From \(2p+q=0\), we get \(q=-2p\). Substitute into the second equation:

\(-2p+4(-2p)=10\Rightarrow -10p=10\Rightarrow p=-1\).

Thus \(q=2\), and

\(x_p=2\sin t-\cos t\).

So the general solution is

\(x=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)+2\sin t-\cos t\).

Now use the initial conditions. When \(t=0\),

\(5=A+0-1\), so \(A=6\).

Differentiate the general solution:

\(\dot{x}=\mathrm{e}^{-t}(-A\cos 2t-B\sin 2t)+\mathrm{e}^{-t}(-2A\sin 2t+2B\cos 2t)+2\cos t+\sin t\).

At \(t=0\), this gives

\(2=-A+2B+2\).

With \(A=6\),

\(2=-6+2B+2\Rightarrow 2B=6\Rightarrow B=3\).

Therefore the particular solution is

\(x=\mathrm{e}^{-t}(6\cos 2t+3\sin 2t)+2\sin t-\cos t\).

As \(t\to \infty\), the exponential term tends to \(0\), so an approximate solution for large positive \(t\) is

\(x\approx 2\sin t-\cos t\).