Answer to Question #311944 in Differential Equations for kxngToooch

Question

Use two different methods to determine L^-1 {(11 - 3s)/ (s² + 2s -3)} Show that your answers are equal.

Answer 1

Method 1 (Using Shifting Property)

$\begin{aligned} L^{-1}\Bigg\{\frac{11-3s}{s^2+2s-3}\Bigg\} &= L^{-1}\Bigg\{\frac{-3(s+1)+14}{s^2+2s+1-4}\Bigg\}\\ &= L^{-1}\Bigg\{\frac{-3(s+1)+14}{(s+1)^2-4}\Bigg\}\\ &= e^{-t} L^{-1}\Bigg\{\frac{-3s+14}{s^2-4}\Bigg\}\\ &= e^{-t} \Bigg[L^{-1}\Big\{\frac{-3s}{s^2-4}\Big\}+L^{-1}\Big\{\frac{14}{s^2-4}\Big\}\Bigg]\\ &= e^{-t} \Big[-3\cosh 2t + 7\sinh 2t\Big]\\ &= e^{-t} \Bigg[-3\Big(\dfrac{e^t + e^{-t}}{2}\Big) + 7\Big(\dfrac{e^t - e^{-t}}{2}\Big)\Bigg]\\ &= \frac{e^{-t}}{2} \Bigg[-3(e^{2t} + e^{-2t}) + 7(e^{2t} - e^{-2t})\Bigg]\\ &= \frac{e^{-t}}{2} \Bigg[4e^{2t} - 10e^{-2t}\Bigg]\\ &= 2e^{t} - 5e^{-3t} \end{aligned}$

Method 2 (Partial fraction method)

We use partial fraction to split $\dfrac{11-3s}{s^2+2s-3}$.

$\begin{aligned} \end{aligned}$ $\begin{aligned} \dfrac{11-3s}{s^2+2s-3} &= \frac{A}{s+3} + \frac{B}{s-1}\\ 11-3s &= A(s-1)+B(s+3)\\ \end{aligned}\\ \text{Taking } s=1 \implies 8 = 4B \implies B=2\\\ \\ \text{Taking } s=-3 \implies 20 = -4A \implies A=-5\\\ \\ \therefore \dfrac{11-3s}{s^2+2s-3} = \dfrac{-5}{s+3} + \dfrac{2}{s-1}\\$

$\begin{aligned} L^{-1} \Bigg\{\dfrac{11-3s}{s^2+2s-3}\Bigg\} &= L^{-1} \Bigg\{\frac{-5}{s+3} + \frac{2}{s-1}\Bigg\}\\ &= 2e^t - 5e^{-3t} \end{aligned}$

From the above two methods, we see that the answers are equal.