Question 4.2.5: Solve y" - 3y' + 2y e^-4, y(0) 1, y'(0) 5....
Solve y” – 3y’ + 2y = e^{-4 t}, y(0) = 1, y'(0) = 5.
Proceeding as in Example 4, we transform the DE by taking the sum of the transforms of each term, use (6) and (7), use the given initial conditions, use part (c) of Theorem 4.1.1, and then solve for Y(s) :
\begin{array}{r} \mathscr{L}\left\{\begin{array}{l} d^{2} y \\\hline d t^{2} \end{array}\right\}-3 \mathscr{L}\left\{\begin{array}{l} d y \\ \hline d t \end{array}\right\}+2 \mathscr{L}\{y\} = \mathscr{L}\left\{e^{-4 t}\right\} \\ s^{2} Y(s)-s y(0)-y^{\prime}(0)-3[s Y(s)-y(0)]+2 Y(s) = \frac{1}{s+4} \end{array}
\begin{aligned} & \left(s^{2}-3 s+2\right) Y(s) = s+2+\frac{1}{s+4} \\ Y(s) = & \frac{s+2}{s^{2}-3 s+2}+\frac{1}{\left(s^{2}-3 s+2\right)(s+4)} = \frac{s^{2}+6 s+9}{(s-1)(s-2)(s+4)}& (14) \end{aligned}
and so y(t) = \mathscr{L}^{-1}\{Y(s)\}. The details of the decomposition of Y(s) into partial fractions have already been carried out in Example 3. In view of (4) and (5) the solution of the initial-value problem is y(t) =-\frac{16}{5} e^{t}+\frac{25}{6} e^{2 t}+\frac{1}{30} e^{-4 t}.