Find the inverse Laplace transform of the function:

[tex]H(s) = \frac{2s}{(s^2 + 1)^2}[/tex]

Is the inverse Laplace transform:

[tex]L^{-1} \left\{ \frac{2s}{(s^2 + 1)^2} \right\} = 2 \cos t + \sin t[/tex]

or

[tex]t \sin t[/tex]?

To find the inverse Laplace transform of a given function, we can often use standard transform tables or known results from Laplace transform theory. Here, we are given a function:
H ( s ) = ( s 2 + 1 ) 2 2 s ​
We are tasked with finding the inverse Laplace transform:
L − 1 { ( s 2 + 1 ) 2 2 s ​ }
In this case, the function involves ( s 2 + 1 ) 2 s ​ , and one should be aware that simplifications and standard inverse Laplace transform results can be utilized. The relevant formula to remember is:

The inverse Laplace transform of s 2 + 1 s ​ is cos t .
The inverse Laplace transform of ( s 2 + 1 ) 2 s ​ is a standard transform and is given by t sin t .

Therefore, applying the known transform, we find:
L − 1 { ( s 2 + 1 ) 2 2 s ​ } = 2 t sin t
Thus, the correct inverse Laplace transform of ( s 2 + 1 ) 2 2 s ​ is:
2 t sin t
This means the inverse Laplace transform provided in the statement 2 cos t + sin t or t sin t is incorrect as stated. The accurate inverse Laplace result is 2 t sin t .

Answered by OliviaMariThompson | 2025-07-06