Please provide steps The inverse Laplace transform for F(s)=8s+9−6s2−3 is a) 8e−9t−6sin⁡h(3t) b) 8e−9t−6cos⁡h(3t) c) 8e9t−6sin⁡h(3t) d) 8e9t−6cos⁡h(3t)

Step 1 Consider the function: F(s)=8x+9−6x2−3 Find Inverse Laplace Transform. Apply Partial fraction on 6x2−3 6x2−3=Ax−34+Bx+34 6=A(x+34)+B(x−34) Let x=34 and x=−34 and find A and B A=334 B=−334 F(s)=8x+9−334(x−34)+334(x+34) Step 2 Inverse Laplace transform property L−1{af(s)+g(s)}=aL−1{f(s)}+L−1{g(s)} and L−11s+a=e−at L−1{8x+9−334(x−34)+334(x+34)}=8L−1{1x+9}−334L−1{1x−34}+334L−1{1x+34} =8e−9t−334e34t+334e−34t

Get the answer to "Please provide steps The inverse Laplace transform for..."

High School
Other

Albarellak

Answered question

2020-10-25

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 & Explanation

liingliing8

Skilled2020-10-26Added 95 answers

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}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Other

Didn't find what you were looking for?