Consider the following second-order differential equation:

y” – 4y’ + 3y = 2t – 3t²

The particular solution of the differential equation is

Explanation:

y” – 4y’ + 3y = 2t – 3t²

f(D) = D² – 4D + 3

\(\begin{array}{l} P.I = \frac{I}{{f\left( D \right)}}\left( {2t - 3{t^2}} \right)\\ = \frac{1}{{{D^2} - 4D + 3}}\left( {2t - 3{t^2}} \right)\\ = \left[ {\frac{1}{{1 - D}} - \frac{1}{{3\left( {1 - \frac{D}{3}} \right)}}} \right]\left( {t - \frac{{3{t^2}}}{2}} \right)\\ \left( {1 + D + {D^2} + \ldots } \right)\left( {t - \frac{{3{t^2}}}{2}} \right) - \frac{1}{3}\left( {t - \frac{D}{3} + \frac{{{D^2}}}{9} + \ldots } \right) \times \left( {t - \frac{{32{t^2}}}{2}} \right)\\ = \left( {t - \frac{{3{t^2}}}{2} + 1 - 3t - 3} \right) - \frac{1}{3}\left( {1 - \frac{{3{t^2}}}{2} + \frac{1}{3} - 8 - \frac{1}{3}} \right) \end{array}\)