Answer: (a) The general solution is
\(x=\mathrm{e}^{-5t}(At+B)+12\sin t-5\cos t\),
where \(A\) and \(B\) are constants.
(b) For large positive values of \(t\), the term \(\mathrm{e}^{-5t}(At+B)\) tends to \(0\), so
\(x\approx 12\sin t-5\cos t=13\sin(t-\phi)\),
with \(R=13\) and
\(\cos\phi=\frac{12}{13},\quad \sin\phi=\frac{5}{13}\),
so \(\phi=\tan^{-1}\!\left(\frac{5}{12}\right)\approx 0.395\text{ rad}\) \((\approx 22.6^\circ)\).
(a) Solve the associated homogeneous equation
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+10\dfrac{\mathrm{d}x}{\mathrm{d}t}+25x=0\).
The auxiliary equation is
\(m^2+10m+25=0=(m+5)^2\),
so there is a repeated root \(m=-5\).
Hence the complementary function is
\(x_c=\mathrm{e}^{-5t}(At+B)\).
For a particular integral, try
\(x_p=p\sin t+q\cos t\).
Then
\(x_p'=p\cos t-q\sin t\),
\(x_p''=-p\sin t-q\cos t\).
Substituting into the differential equation gives
\((-p\sin t-q\cos t)+10(p\cos t-q\sin t)+25(p\sin t+q\cos t)=338\sin t\).
Collect coefficients:
\((24p-10q)\sin t+(10p+24q)\cos t=338\sin t\).
So
\(24p-10q=338\),
\(10p+24q=0\).
From \(10p+24q=0\), we have \(5p+12q=0\), so \(p=-\dfrac{12}{5}q\).
Substitute into the first equation:
\(24\left(-\dfrac{12}{5}q\right)-10q=338\).
Multiplying by 5:
\(-288q-50q=1690\),
\(-338q=1690\),
so \(q=-5\).
Then \(10p+24(-5)=0\), so \(10p=120\) and \(p=12\).
Thus
\(x_p=12\sin t-5\cos t\).
Therefore the general solution is
\(x=x_c+x_p=\mathrm{e}^{-5t}(At+B)+12\sin t-5\cos t\).
(b) For large positive \(t\), the factor \(\mathrm{e}^{-5t}\to 0\), so for any initial conditions
\(x\approx 12\sin t-5\cos t\).
Write this as \(R\sin(t-\phi)\). Since
\(R\sin(t-\phi)=R\sin t\cos\phi-R\cos t\sin\phi\),
comparing coefficients with \(12\sin t-5\cos t\) gives
\(R\cos\phi=12\),
\(R\sin\phi=5\).
Squaring and adding,
\(R^2=12^2+5^2=169\),
so \(R=13\).
Then
\(\cos\phi=\dfrac{12}{13},\quad \sin\phi=\dfrac{5}{13}\),
hence
\(\phi=\tan^{-1}\!\left(\dfrac{5}{12}\right)\approx 0.395\text{ rad}\).
So for large positive \(t\),
\(x\approx 13\sin(t-0.395)\).
