Example 5.5.1.

Find the general solution of the system \begin {align*} Dy + y -3x & = t\hspace {0.4cm} \text {where}\hspace {0.4cm} D = \frac {d}{dt}\hspace {0.1cm}, \hspace {0.3cm} D^2 = \frac {d^2}{dt^2}\hspace {0.1cm}, \hspace {0.3cm} \frac {1}{D}= \int dt\hspace {0.1cm}, \hspace {0.3cm} Dy = y'\hspace {0.1cm}\\ \hspace {0.3cm} Dx &= x' Dx + 2x - 2y = -t \end {align*}

\begin {align*} \big (D + 1\big ) y - 3x & = t\\ -2y + \big (D + 2\big )x & = -t \end {align*}

Solving the equations simultaneously \begin {align*} 2\big (D + 1\big ) - 6x & = 2t\\ -2\big (D + 1\big )y + \big (D + 2\big )\big (D + 1\big ) & = -\big (D + 1\big ) t\\\\ \implies \hspace {1cm} \Big [ \big (D + 1\big )\big (D + 2\big ) - 6\Big ] x & = 2t - \big (1 + t\big ) \end {align*}

\[\big (D^2 + 3D - 4\big ) x = t -1 \]

Solving \(\big (D^2 + 3D - 4\big )x = 0\)

\[D^2 + 3D - 4 = 0\hspace {0.4cm}, \hspace {0.5cm} D = 1,-4\]

\[ x_c = A e^{t} + B e^{-4t}\]

\begin {align*} \implies x_p & = \frac {1}{D^2 + 3D - 4} \hspace {0.1cm} (t -1) = \frac {1}{\big (D - 1\big )\big (D + 4\big )}\hspace {0.1cm}(t-1)\\\\ & = \frac {-1}{\big (1 - D\big )}\cdot \frac {1}{4\big (1 + D/4\big )}\hspace {0.1cm}(t-1)\\\\ & = \frac {-1}{4}\hspace {0.1cm} \big (1 - D\big )^{-1} \hspace {0.1cm}\big (1 + D/4\big )^{-1}\hspace {0.1cm}(t-1)\\ \end {align*}

Now we use the binomial theorem to expand \(\big (1 - D\big )^{-1}\) and \(\big (1 + D/4\big )^{-1}\)

\begin {align*} \big (1 - D\big )^{-1} & = \big (1 + D + D^2 + \cdots \cdots \cdots \big )\\\\ (1 + D/4\big )^{-1} & = \big (1 - D/4 + D^2/16 - + \cdots \cdots \cdots \big )\hspace {0.5cm} \text {But}\hspace {0.3cm} D^2(t-1) = 0 \end {align*}

\begin {align*} \implies \hspace {0.5cm} x_p & = \frac {-1}{4}\hspace {0.1cm} \big (1 - D\big )^{-1} \hspace {0.1cm}\big (1 + D/4\big )^{-1}\hspace {0.1cm}(t-1)\\ & = \frac {-1}{4}\hspace {0.1cm}\big (1 + D + \cdots \cdots \big ) \big (1 - D/4 + \cdots \cdots \big ) (t -1)\\ & = \frac {-1}{4}\hspace {0.1cm}\bigg ( 1 - \frac {D}{4} + D + \cdots \cdots \bigg ) (t-1)\\ & = \frac {-1}{4}\hspace {0.1cm}\Big [ (t-1) - \frac {D}{4}(t -1) + D (t-1)\Big ]\\ & = \frac {-t}{4} + \frac {1}{4} + \frac {1}{16} - \frac {1}{4}\\ & = \frac {1}{4}\Big (\frac {1}{4}-t\Big ) \end {align*}

\[\implies \hspace {1cm} x(t) = x_c + x_p\]

\[x(t) = Ae^t + Be^{-4t} + \frac {1}{4}\Big (\frac {1}{4} - t\Big )\]

\begin {align*} -2y & = -t -Dx -2x\\ 2y & = t +Dx + 2x \end {align*}

\[\implies \hspace {0.5cm} 2y = t + Ae^t -4Be^{-4t} - \frac {1}{4} + 2Ae^t + 2Be^{-4t} +\frac {1}{2}\Big (\frac {1}{4}-t\Big ) \]

\[\implies \hspace {0.3cm} 2y = 3Ae^t - 2Be^{-4t} + \frac {t}{2} - \frac {1}{8}\]

\[\implies \hspace {1cm} y(t) = \frac {3}{2}Ae^t - B e^{-4t} + \frac {t}{4} - \frac {1}{16}\]