Answer: (a) The general solution is \(y=A\mathrm{e}^{-x}+B\mathrm{e}^{-2x}+x-1\).

(b) For large positive values of \(x\), an approximate solution is \(y\approx x-1\).

(a) First solve the homogeneous equation

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+3\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y=0\).

Using a trial solution \(y=\mathrm{e}^{mx}\), the auxiliary equation is

\(m^2+3m+2=0\).

So

\((m+1)(m+2)=0\), giving \(m=-1\) and \(m=-2\).

Hence the complementary function is

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

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

\(y_p=p+qx\).

Then

\(y_p'=q, \quad y_p''=0\).

Substituting into the differential equation gives

\(0+3q+2(p+qx)=2x+1\).

So

\(2qx+(3q+2p)=2x+1\).

Equating coefficients:

\(2q=2\), so \(q=1\),

and \(3q+2p=1\), so \(3+2p=1\), hence \(p=-1\).

Therefore

\(y_p=x-1\).

So the general solution is

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

(b) As \(x\to +\infty\), \(\mathrm{e}^{-x}\to 0\) and \(\mathrm{e}^{-2x}\to 0\).

Therefore, for large positive values of \(x\),

\(y\approx x-1\).