Solution. In operator form, \[(D + 1) y - 3x = t ,\qquad -2y + (D + 2)x = -t .\] Multiply the first by \(2\) and apply \(D + 1\) to the second: \[2(D + 1)y - 6x = 2t ,\qquad -2(D + 1)y + (D + 2)(D + 1)x = -(D + 1)t = -(1 + t) .\] Adding eliminates \(y\): \[\big [(D + 1)(D + 2) - 6\big ]x = 2t - (1 + t),\qquad \text {that is}\qquad (D^2 + 3D - 4) x = t - 1 .\] The characteristic equation \(m^2 + 3m - 4 = 0\) has roots \(1\) and \(-4\), so \(x_c = A e^{t} + B e^{-4t}\). For a particular solution, \[x_p = \frac {1}{(D - 1)(D + 4)}\,(t-1) = -\frac {1}{4}(1 - D)^{-1}\left (1 + \frac {D}{4}\right )^{-1}(t-1) .\] Expanding by the binomial series, \((1 - D)^{-1} = 1 + D + D^2 + \cdots \) and \(\left (1 + \frac {D}{4}\right )^{-1} = 1 - \frac {D}{4} + \cdots \), and since \(D^2(t-1) = 0\) only the terms up to \(D\) matter: \[x_p = -\frac {1}{4}\left ( 1 + D - \frac {D}{4}\right ) (t-1) = -\frac {1}{4}\left [(t-1) + \frac 34\right ] = -\frac {t}{4} + \frac {1}{16} .\] So \[x(t) = Ae^t + Be^{-4t} - \frac {t}{4} + \frac {1}{16} .\] The second equation gives \(2y = Dx + 2x + t\), so \[2y = \left (Ae^t - 4Be^{-4t} - \frac 14\right ) + \left (2Ae^t + 2Be^{-4t} - \frac {t}{2} + \frac 18\right ) + t = 3Ae^t - 2Be^{-4t} + \frac {t}{2} - \frac {1}{8} ,\] and \[y(t) = \frac {3}{2}Ae^t - B e^{-4t} + \frac {t}{4} - \frac {1}{16} .\]

Solution. For an exponential, \(P(D)e^{at} = P(a)e^{at}\), so \(\frac {1}{P(D)}e^{at} = \frac {e^{at}}{P(a)}\) when \(P(a)\neq 0\). Here \(P(3) = 9 + 9 + 2 = 20\), so \(x_p = \frac {4e^{3t}}{20} = \frac 15 e^{3t}\).