Answer: \(x=e^{5t}+2e^{-2t}+0.3\cos t-0.1\sin t\).

We solve the homogeneous equation first:

\(\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}-3\dfrac{\mathrm{d}x}{\mathrm{d}t}-10x=0\).

Trying \(x=e^{mt}\) gives the auxiliary equation

\(m^2-3m-10=0\), so \((m-5)(m+2)=0\).

Hence the complementary function is

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

For a particular solution, because the forcing term is \(2\sin t-3\cos t\), let

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

Then

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

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

Substitute into the differential equation:

\(-p\sin t-q\cos t-3(p\cos t-q\sin t)-10(p\sin t+q\cos t)=2\sin t-3\cos t\).

Collecting coefficients of \(\sin t\) and \(\cos t\),

\((-11p+3q)\sin t+(-3p-11q)\cos t=2\sin t-3\cos t\).

So

\(-11p+3q=2\), \(3p+11q=3\).

Solving these gives \(p=-0.1\) and \(q=0.3\).

Therefore

\(x=Ae^{5t}+Be^{-2t}+0.3\cos t-0.1\sin t\).

Now use the initial conditions. When \(t=0\), \(x=3.3\), so

\(A+B+0.3=3.3\), hence \(A+B=3\).

Differentiate:

\(\dfrac{\mathrm{d}x}{\mathrm{d}t}=5Ae^{5t}-2Be^{-2t}-0.3\sin t-0.1\cos t\).

When \(t=0\), \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=0.9\), so

\(5A-2B-0.1=0.9\), hence \(5A-2B=1\).

Solve

\(A+B=3\), \(5A-2B=1\).

From \(B=3-A\),

\(5A-2(3-A)=1 \Rightarrow 7A=7 \Rightarrow A=1\), and then \(B=2\).

So the required solution is

\(x=e^{5t}+2e^{-2t}+0.3\cos t-0.1\sin t\).