Final answer:
The Inverse Laplace transform of 3s+4/s²+4s+85 is found by completing the square in the denominator and using standard transform tables to find that it is 3e^-2t*cos(9t) + (4/9)e^-2t*sin(9t).
Step-by-step explanation:
The question is asking to find the Inverse Laplace transform of the function 3s+4/s²+4s+85. To approach this problem, we first rewrite the denominator to complete the square, and make it more apparent how to decompose the function into a form that is recognizable from standard Laplace transform tables.
We have:
s² + 4s + 85 = (s + 2)² + 81
Now, we can write our function as:
(3s+4) / ((s+2)² + 81)
To simplify further, we separate this into two parts, one with 's' in the numerator and the other with a constant in the numerator, which will allow us to use known inverse transforms:
A = 3s / ((s+2)² + 81)
B = 4 / ((s+2)² + 81)
The inverse Laplace transform for A uses the form which is the transform of a cosine function, and for B, it uses the transform of a sine function:
L-1{A} = 3e-2tcos(9t) and L-1{B} = (4/9)e-2tsin(9t)
Therefore, the inverse Laplace transform of the original function is:
L-1{(3s+4)/((s+2)² + 81)}
= 3e-2tcos(9t) + (4/9)e-2tsin(9t)