Answer: The general solution is \(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).
First solve the complementary equation
\(\dfrac{\mathrm d^2x}{\mathrm dt^2}-6\dfrac{\mathrm dx}{\mathrm dt}+25x=0\).
Its auxiliary equation is \(m^2-6m+25=0\), so
\(m=\dfrac{6\pm \sqrt{36-100}}{2}=3\pm 4\mathrm i\).
Hence the complementary function is
\(x_c=e^{3t}(A\cos 4t+B\sin 4t)\).
Now seek a particular integral of the form
\(x_p=p\sin 2t+q\cos 2t\).
Then
\(\dfrac{\mathrm dx_p}{\mathrm dt}=2p\cos 2t-2q\sin 2t\),
\(\dfrac{\mathrm d^2x_p}{\mathrm dt^2}=-4p\sin 2t-4q\cos 2t\).
Substitute into the differential equation:
\(\left(-4p\sin 2t-4q\cos 2t\right)-6\left(2p\cos 2t-2q\sin 2t\right)+25\left(p\sin 2t+q\cos 2t\right)=195\sin 2t\).
Collect coefficients of \(\sin 2t\) and \(\cos 2t\):
\(\left(21p+12q\right)\sin 2t+\left(-12p+21q\right)\cos 2t=195\sin 2t\).
So
\(21p+12q=195\), \(\,-12p+21q=0\).
From \(-12p+21q=0\), we get \(p=\dfrac{7}{4}q\). Substituting into the first equation gives
\(21\cdot \dfrac{7}{4}q+12q=195\), hence \(\dfrac{195}{4}q=195\), so \(q=4\) and \(p=7\).
Therefore
\(x_p=7\sin 2t+4\cos 2t\).
Combining the complementary function and a particular integral gives the general solution:
\(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).
