Please provide steps The inverse Laplace transform for F(s)=8/(s+9)-6/(s^2-sqrt3)

Question

Please provide steps
The inverse Laplace transform for
$F (s) = \frac{8}{s + 9} - \frac{6}{s^{2} - \sqrt{3}}$ is
a) $8 e^{- 9 t} - 6 \sin h (3 t)$
b) $8 e^{- 9 t} - 6 \cos h (3 t)$
c) $8 e^{9 t} - 6 \sin h (3 t)$
d) $8 e^{9 t} - 6 \cos h (3 t)$

Answer 1

Step 1
Consider the function:

F (s) = \frac{8}{x + 9} - \frac{6}{x^{2} - \sqrt{3}}

Find Inverse Laplace Transform.
Apply Partial fraction on

\frac{6}{x^{2} - \sqrt{3}}

\frac{6}{x^{2} - \sqrt{3}} = \frac{A}{x - \sqrt[4]{3}} + \frac{B}{x + \sqrt[4]{3}}

6 = A (x + \sqrt[4]{3}) + B (x - \sqrt[4]{3})

Let

x = \sqrt[4]{3} and x = - \sqrt[4]{3}

and find A and B

A = \frac{3}{\sqrt[4]{3}}

B = - \frac{3}{\sqrt[4]{3}}

F (s) = \frac{8}{x + 9} - \frac{3}{\sqrt[4]{3} (x - \sqrt[4]{3})} + \frac{3}{\sqrt[4]{3} (x + \sqrt[4]{3})}

Step 2
Inverse Laplace transform property

L^{- 1} {a f (s) + g (s)} = a L^{- 1} {f (s)} + L^{- 1} {g (s)} and L^{- 1} 1 s + a = e^{- a t}

L^{- 1} {\frac{8}{x + 9} - \frac{3}{\sqrt[4]{3} (x - \sqrt[4]{3})} + \frac{3}{\sqrt[4]{3} (x + \sqrt[4]{3})}} = 8 L^{- 1} {\frac{1}{x + 9}} - 3^{\frac{3}{4}} L^{- 1} {\frac{1}{x - \sqrt[4]{3}}} + 3^{\frac{3}{4}} L^{- 1} {\frac{1}{x + \sqrt[4]{3}}}

= 8 e^{- 9 t} - 3^{\frac{3}{4}} e^{\sqrt[4]{3} t} + 3^{\frac{3}{4}} e^{- \sqrt[4]{3} t}

Please provide steps The inverse Laplace transform for F(s)=8/(s+9)-6/(s^2-sqrt3)

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