Answer :
To find the inverse Laplace transform of [tex]\(\frac{4s}{4s^2 + 1}\)[/tex], we can use standard techniques and known transforms. Here's a detailed, step-by-step solution:
Given:
[tex]\[ L^{-1}\left(\frac{4s}{4s^2 + 1}\right) \][/tex]
First, let's simplify the given expression. Notice that the denominator can be factored as follows:
[tex]\[ 4s^2 + 1 = \left(2s\right)^2 + 1 \][/tex]
Thus, the given fraction can be rewritten as:
[tex]\[ \frac{4s}{4s^2 + 1} = \frac{4s}{(2s)^2 + 1} \][/tex]
Now, let's factor out constants to further simplify this:
[tex]\[ \frac{4s}{(2s)^2 + 1} = \frac{4 \cdot s}{4 \cdot (\left(\frac{2s}{2}\right)^2 + 1)} = \frac{4s}{4((\frac{2s}{2})^2 + 1)} = \frac{4s}{4} \cdot \frac{1}{\left(\frac{2s}{2}\right)^2 + 1} = s \cdot \frac{1}{\left(\frac{2s}{2}\right)^2 + 1} \][/tex]
Therefore, we simplify to:
[tex]\[ \frac{4s}{4s^2 + 1} = s \cdot \frac{1}{(2s)^2 + 1} \][/tex]
We now recognize that this is in the form [tex]\( s \cdot \frac{1}{s^2 + \alpha^2} \)[/tex], where [tex]\(\alpha = 1/2\)[/tex].
Recall the inverse Laplace transform for the standard form:
[tex]\[ L^{-1}\left(\frac{s}{s^2 + \alpha^2}\right) = \cos(\alpha t) \][/tex]
In our case, [tex]\(\alpha = \frac{1}{2}\)[/tex], so we have:
[tex]\[ L^{-1}\left(\frac{s}{s^2 + \left(\frac{1}{2}\right)^2}\right) = \cos\left(\frac{1}{2}t\right) \][/tex]
So, the inverse Laplace transform is:
[tex]\[ L^{-1}\left(\frac{4s}{4s^2 + 1}\right) = \cos\left(\frac{1}{2}t\right) \][/tex]
Thus, the solution is:
[tex]\[ \boxed{\cos\left(\frac{1}{2}t\right)} \][/tex]
Given:
[tex]\[ L^{-1}\left(\frac{4s}{4s^2 + 1}\right) \][/tex]
First, let's simplify the given expression. Notice that the denominator can be factored as follows:
[tex]\[ 4s^2 + 1 = \left(2s\right)^2 + 1 \][/tex]
Thus, the given fraction can be rewritten as:
[tex]\[ \frac{4s}{4s^2 + 1} = \frac{4s}{(2s)^2 + 1} \][/tex]
Now, let's factor out constants to further simplify this:
[tex]\[ \frac{4s}{(2s)^2 + 1} = \frac{4 \cdot s}{4 \cdot (\left(\frac{2s}{2}\right)^2 + 1)} = \frac{4s}{4((\frac{2s}{2})^2 + 1)} = \frac{4s}{4} \cdot \frac{1}{\left(\frac{2s}{2}\right)^2 + 1} = s \cdot \frac{1}{\left(\frac{2s}{2}\right)^2 + 1} \][/tex]
Therefore, we simplify to:
[tex]\[ \frac{4s}{4s^2 + 1} = s \cdot \frac{1}{(2s)^2 + 1} \][/tex]
We now recognize that this is in the form [tex]\( s \cdot \frac{1}{s^2 + \alpha^2} \)[/tex], where [tex]\(\alpha = 1/2\)[/tex].
Recall the inverse Laplace transform for the standard form:
[tex]\[ L^{-1}\left(\frac{s}{s^2 + \alpha^2}\right) = \cos(\alpha t) \][/tex]
In our case, [tex]\(\alpha = \frac{1}{2}\)[/tex], so we have:
[tex]\[ L^{-1}\left(\frac{s}{s^2 + \left(\frac{1}{2}\right)^2}\right) = \cos\left(\frac{1}{2}t\right) \][/tex]
So, the inverse Laplace transform is:
[tex]\[ L^{-1}\left(\frac{4s}{4s^2 + 1}\right) = \cos\left(\frac{1}{2}t\right) \][/tex]
Thus, the solution is:
[tex]\[ \boxed{\cos\left(\frac{1}{2}t\right)} \][/tex]
