Answer: (a) \(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}+t^2+1\), where \(A\) and \(B\) are constants.

(b) For large positive values of \(t\), \(x\approx t^2+1\).

(a) First solve the homogeneous equation

\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+5\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=0\).

Using \(x=\mathrm{e}^{mt}\), the auxiliary equation is

\(6m^2+5m+1=0\).

This factorises as \((3m+1)(2m+1)=0\), so

\(m=-\dfrac12\) or \(m=-\dfrac13\).

Hence the complementary function is

\(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}\).

For a particular integral, since the right-hand side is quadratic, try

\(x=p+qt+rt^2\).

Then

\(x'=q+2rt\), \(x''=2r\).

Substitute into the differential equation:

\(6(2r)+5(q+2rt)+(p+qt+rt^2)=t^2+10t+13\).

So

\(12r+5q+10rt+p+qt+rt^2=t^2+10t+13\).

Equating coefficients gives

\(r=1\), \(q=0\), \(p=1\).

So the particular integral is \(t^2+1\).

Therefore the general solution is

\(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}+t^2+1\).

(b) For large positive values of \(t\), the exponential terms become negligible, so

\(x\approx t^2+1\).