Final answer:
The inverse Laplace transform of F(s) = (1/s-3) - (16/s²+9) is f(t) = e³t - 16sin(3t), obtained by taking the inverse transforms of each term and using the linearity property.
Step-by-step explanation:
To find the inverse Laplace transform of F(s) = (1/s-3) - (16/s²+9), we can break it down into simpler parts that correspond to known Laplace transforms.
Step 1: Simplify the expression
Firstly, we express F(s) in a form where each term corresponds to a standard Laplace transform.
We have:
1/(s - 3), which corresponds to an exponential shift,
16/(s² + 9), which is the transform of a sine function,
Step 2: Apply the inverse Laplace transform
Using the linearity of the Laplace transform and known inverse transforms, we get:
The inverse Laplace transform of 1/(s - 3) is e³t.
The inverse Laplace transform of 16/(s² + 9) is 16sin(3t), since the coefficient of t in the sine function is the square root of 9, which is 3.
Combining both results using linearity:
The inverse Laplace transform of F(s) is f(t) = e³t - 16sin(3t).