Question

35 Determine the inverse Laplace transform of: 1 2 32-8

Answer 1

Key points:

• $L^{-1}\left \{ \frac{1}{s-a} \right \}=e^{at}$
• 
  S (-1 = cosh(at) s2 - a2
  

We have,

L-1 3s 152 -8 L-1 6s $2 – 16

$=L^{-1}\left \{ \frac{6s}{(s-4)(s+4)} \right \}$

$=6L^{-1}\left \{ \frac{s}{(s-4)(s+4)} \right \}$ [Using linearity of inverse laplace transform]

$=\frac{6}{2}L^{-1}\left \{ \frac{1}{s-4}+\frac{1}{s+4} \right \}$

$=3L^{-1}\left \{ \frac{1}{s-4}+\frac{1}{s+4} \right \}$

$=3L^{-1}\left \{ \frac{1}{s-4}\right \}+3L^{-1}\left \{\frac{1}{s+4} \right \}$  [Using linearity of inverse laplace transform]

$=3e^{4t}+3e^{-4t}$ [Using key points]

$\therefore {\color{Blue} L^{-1}\left \{ \frac{3s}{\frac{1}{2}s^2-8} \right \}=3e^{4t}+3e^{-4t}}$

Alternatively,

L-1 3s 152 -8 L-1 6s $2 – 16

$=6L^{-1}\left \{ \frac{s}{s^2-4^2} \right \}$ [Using linearity of inverse laplace transform]

$=6\cosh(4t)$

$\therefore {\color{Blue} L^{-1}\left \{ \frac{3s}{\frac{1}{2}s^2-8} \right \}=6\cosh(4t)}$

Hope this helps!