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

(ii) For large positive \(t\), the exponential terms decay to zero, so an approximate solution is \(x\approx -2\cos t\).

We solve the differential equation \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+2\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=4\sin t\).

First consider the associated homogeneous equation

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

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

\(m^2+2m+1=0\), so \((m+1)^2=0\) and hence \(m=-1\) is a repeated root.

Therefore the complementary function is

\(x_h=(A+Bt)e^{-t}\),

where \(A\) and \(B\) are constants.

Now find a particular integral. Since the forcing term is \(4\sin t\), try

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

Then

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

and

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

Substitute into the differential equation:

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

The \(\sin t\) terms give \(-2q=4\), so \(q=-2\). The \(\cos t\) terms give \(2p=0\), so \(p=0\).

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

So the general solution is

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

For large positive \(t\), the term \((A+Bt)e^{-t}\) tends to \(0\), so an approximate solution is

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