Question 4.2.3: Partial Fractions and Linearity Evaluate L^-1{ s² + 6s +9/(......
Partial Fractions and Linearity
Evaluate \mathscr{L}^{-1}\left\{\left(s-\frac{s^{2}+6 s+9}{1)(s-2)(s+4)}\right\}\right..
There exist unique constants A, B, and C such that
\frac{s^{2}+6 s+9}{(s-1)(s-2)(s+4)} \quad-\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s+4}\frac{A(s-2)(s+4)+B(s-1)(s+4)+C(s-1)(s-2)}{(s-1)(s-2)(s+4)}
Since the denominators are identical, the numerators are identical:
s^{2}+6 s+9 \quad A(s-2)(s+4)+B(s-1)(s+4)+C(s-1)(s-2) . (3)
By comparing coefficients of powers of s on both sides of the equality, we know that (3) is equivalent to a system of three equations in the three unknowns A, B, and C. However, recall that there is a shortcut for determining these unknowns. If we set s \quad 1, s \quad 2, and s=-4 in (3) we obtain, respectively,*
16 \quad A(-1)(5), 25 \quad B(1)(6), 1 \quad C(-5)(-6),
and so A=-\frac{16}{5}, B \quad \frac{25}{6}, C \quad \frac{1}{30}. Hence the partial fraction decomposition is
\frac{s^2+6 s+9}{(s-1)(s-2)(s+4)}=-\frac{16 / 5}{s-1}+\frac{25 / 6}{s-2}+\frac{1 / 30}{s+4}, (4)
and thus, from the linearity of \mathscr{L}^{-1} and part (c) of Theorem 4.2.1,
\begin{aligned} \mathscr{L}^{-1}\left\{\frac{s^2+6 s+9}{(s-1)(s-2)(s+4)}\right\} & =-\frac{16}{5} \mathscr{L}^{-1}\left\{\frac{1}{s-1}\right\}+\frac{25}{6} \mathscr{L}^{-1}\left\{\frac{1}{s-2}\right\}+\frac{1}{30} \mathscr{L}^{-1}\left\{\frac{1}{s+4}\right\} \\ & =-\frac{16}{5} e^t+\frac{25}{6} e^{2 t}+\frac{1}{30} e^{-4 t}. &(5) \end{aligned}
*The numbers 1, 2, and -4 are the zeros of the common denominator (s – 1)(s – 2)(s + 4).