Question

Find the inverse Laplace transform L−1{F(s)} of the given function.

F(s) = 8s^2 − 12s + 144 / s(s^2+36)


Your answer should be a function of t. Enclose arguments of functions in parentheses. For example, sin(2x).

L−1{F(s)}= ___________-

Answer 1

Using partial fractions for $F(s) = \dfrac{8s^{2}-12s+144}{s(s^{2}+36)}$ ,

$\begin{aligned} \dfrac{8s^{2}-12s+144}{s(s^{2}+36)} &= \dfrac{A}{s}+\dfrac{Bs+C}{s^{2}+36}\\ 8s^{2}-12s+144 & = A(s^{2}+36) + (Bs+C)s\\ 8s^{2}-12s+144 & =(A+B)s^{2}+Cs+36A \end{aligned}$

Comparing the coefficients of $s^{2}, s$ and constant terms, we get

$\begin{aligned} A+B &= 8\\ C &= -12\\ A = 4 &\Rightarrow B = 4 \end{aligned}$

Therefore,

$\begin{aligned} L^{-1}\{F(s)\} &= L^{-1}\left\{\dfrac{4}{s} + \dfrac{4s-12}{s^{2}+36} \right\}\\ &=L^{-1}\left\{\dfrac{4}{s}\right\}+L^{-1}\left\{\dfrac{4s}{s^{2}+36}\right\}-L^{-1}\left\{\dfrac{12}{s^{2}+36}\right\}\\ &=4L^{-1}\left\{\dfrac{1}{s}\right\}+4L^{-1}\left\{\dfrac{s}{s^{2}+36}\right\}-2L^{-1}\left\{\dfrac{6}{s^{2}+36}\right\}\\ L^{-1}\{F(s)\}&=4+4\cos(6t)-2\sin(6t) \end{aligned}$