Answer: (a) \(x=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt{3}}{2}t+B\cos \frac{\sqrt{3}}{2}t\right)+t^2-2t+1\).
(b) For large positive values of \(t\), \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).
(a) First solve the homogeneous equation
\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+\frac{\mathrm{d}x}{\mathrm{d}t}+x=0\).
The auxiliary equation is
\(m^2+m+1=0\).
So
\(m=\frac{-1\pm\sqrt{1-4}}{2}=-\frac12\pm i\frac{\sqrt3}{2}\).
Hence the complementary function is
\(x_h=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt3}{2}t+B\cos \frac{\sqrt3}{2}t\right)\).
For a particular solution, since the right-hand side is \(t^2+1\), try
\(x_p=pt^2+qt+r\).
Then
\(\frac{\mathrm{d}x_p}{\mathrm{d}t}=2pt+q\), \(\frac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=2p\).
Substituting into the differential equation gives
\(2p+(2pt+q)+(pt^2+qt+r)=t^2+1\).
So
\(pt^2+(2p+q)t+(2p+q+r)=t^2+1\).
Equating coefficients:
\(p=1\), \(2p+q=0\Rightarrow q=-2\), and \(2p+q+r=1\Rightarrow r=1\).
Thus
\(x_p=t^2-2t+1\).
Therefore the general solution is
\(x=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt3}{2}t+B\cos \frac{\sqrt3}{2}t\right)+t^2-2t+1\).
(b) As \(t\to\infty\), the factor \(\mathrm{e}^{-t/2}\to 0\), so the exponential term and its derivatives become negligible.
Hence for large \(t\),
\(x\approx t^2-2t+1\).
Differentiating twice,
\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).
So for large positive values of \(t\), \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).
