Answer :
The inverse Laplace transform of (6 - 2s) / ([tex]s^2[/tex] + 4s + 8) is [tex]\[ 10e^{-2t}\sin(2t) - 2e^{-2t}\cos(2t)\][/tex]
To find the inverse Laplace transform of the given function, we need to express the function in a form that can be easily inverted.
Let's start with the given function:
[tex]\[F(s) = \frac{6 - 2s}{s^2 + 4s + 8}\][/tex]
First, we simplify the denominator:
[tex]\[s^2 + 4s + 8 = (s + 2)^2 + 4\][/tex]
So, the function becomes:
[tex]\[F(s) = \frac{6 - 2s}{(s + 2)^2 + 4}\][/tex]
Next, we can separate the numerator:
[tex]\[F(s) = \frac{6}{(s + 2)^2 + 4} - \frac{2s}{(s + 2)^2 + 4}\][/tex]
To simplify the terms separately, let's start with the first term:
[tex]\[\frac{6}{(s + 2)^2 + 4}\][/tex]
We recognize this as a form of the standard Laplace transform of [tex]\( \frac{A}{s^2 + a^2} \),[/tex] which corresponds to [tex]\( \frac{A}{a} \sin(at) \)[/tex].
Here A = 6 , a = 2, and (s+2) is a shift. Thus, we have:
[tex]\[\mathcal{L}^{-1}\left\{\frac{6}{(s + 2)^2 + 4}\right\} = 6e^{-2t}\sin(2t)\][/tex]
Now, for the second term:
[tex]\[-\frac{2s}{(s + 2)^2 + 4}\][/tex]
We need to manipulate this to fit a recognizable form. Notice that s can be rewritten as (s + 2 - 2) :
[tex]\[-\frac{2s}{(s + 2)^2 + 4} = -\frac{2(s + 2 - 2)}{(s + 2)^2 + 4} = -\frac{2(s + 2)}{(s + 2)^2 + 4} + \frac{4}{(s + 2)^2 + 4}\][/tex]
Now we split this into two parts:
[tex]\[-\frac{2(s + 2)}{(s + 2)^2 + 4} + \frac{4}{(s + 2)^2 + 4}\][/tex]
For the first part:
[tex]\[-\frac{2(s + 2)}{(s + 2)^2 + 4}\][/tex]
We recognize this as a form of the standard Laplace transform of [tex]\( \frac{s}{s^2 + a^2} \)[/tex], which corresponds to [tex]\( \cos(at) \).[/tex]
Thus:
[tex]\[\mathcal{L}^{-1}\left\{-\frac{2(s + 2)}{(s + 2)^2 + 4}\right\} = -2e^{-2t}\cos(2t)\][/tex]
For the second part:
[tex]\[\frac{4}{(s + 2)^2 + 4}\][/tex]
We recognize this as the same form as the first term we calculated:
[tex]\[\mathcal{L}^{-1}\left\{\frac{4}{(s + 2)^2 + 4}\right\} = 4e^{-2t}\sin(2t)\][/tex]
Now, combining all the parts:
[tex]\[F(s) = \frac{6}{(s + 2)^2 + 4} - \frac{2(s + 2)}{(s + 2)^2 + 4} + \frac{4}{(s + 2)^2 + 4}\][/tex]
So the inverse Laplace transform is:
[tex]\[\mathcal{L}^{-1}\{F(s)\} = 6e^{-2t}\sin(2t) - 2e^{-2t}\cos(2t) + 4e^{-2t}\sin(2t)\][/tex]
Combining like terms:
[tex]\[\mathcal{L}^{-1}\{F(s)\} = (6e^{-2t}\sin(2t) + 4e^{-2t}\sin(2t)) - 2e^{-2t}\cos(2t)\][/tex]
[tex]\[= 10e^{-2t}\sin(2t) - 2e^{-2t}\cos(2t)\][/tex]
Thus, the inverse Laplace transform of the given function is:
[tex]\[\boxed{10e^{-2t}\sin(2t) - 2e^{-2t}\cos(2t)}\][/tex]
