Answer: The general solution is \(x=Ae^{-2t}+Be^{-5t}+3\sin 2t-7\cos 2t\).

For large positive values of \(t\), the exponential terms decay to zero, so \(x\approx 3\sin 2t-7\cos 2t\).

First solve the associated homogeneous equation

\(\dfrac{\mathrm d^2x}{\mathrm dt^2}+7\dfrac{\mathrm dx}{\mathrm dt}+10x=0.\)

Try \(x=e^{mt}\). Then \(m^2+7m+10=0\), so

\((m+2)(m+5)=0\), hence \(m=-2\) or \(m=-5\).

So the complementary function is

\(x_h=Ae^{-2t}+Be^{-5t}.\)

Now seek a particular solution in the form

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

Then

\(\dfrac{\mathrm dx_p}{\mathrm dt}=2p\cos 2t-2q\sin 2t,\)

and

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

Substitute into the differential equation:

\(-4p\sin 2t-4q\cos 2t+7(2p\cos 2t-2q\sin 2t)+10(p\sin 2t+q\cos 2t)=116\sin 2t.\)

Collect coefficients of \(\sin 2t\) and \(\cos 2t\):

\((6p-14q)\sin 2t+(14p+6q)\cos 2t=116\sin 2t.\)

Hence

\(6p-14q=116,\qquad 14p+6q=0.\)

From \(14p+6q=0\), divide by 2 to get \(7p+3q=0\), so \(q=-\dfrac{7}{3}p\). Substituting into \(6p-14q=116\):

\(6p-14\left(-\dfrac{7}{3}p\right)=116\)

\(6p+\dfrac{98}{3}p=116\)

\(\dfrac{116}{3}p=116\)

so \(p=3\), and then \(q=-7\).

Therefore \(x_p=3\sin 2t-7\cos 2t\).

So the general solution is

\(x=Ae^{-2t}+Be^{-5t}+3\sin 2t-7\cos 2t.\)

As \(t\to\infty\), the exponential terms vanish, so an approximate solution is

\(x\approx 3\sin 2t-7\cos 2t.\)