Final answer:
To find the inverse Laplace transform of F(s) = 1/(s(s - 3)), we can use Theorem 2 which states that if F(s) is a rational function, then the inverse Laplace transform is given by the sum of partial fraction terms . The inverse Laplace transform ofF(s) is (-1/3) + (1/3)e^(3t).
Step-by-step explanation:
Inverse Laplace transform of F(s) = 1/(s(s - 3))
To find the inverse Laplace transform of F(s), we can use Theorem 2 which states that if F(s) is a rational function of the form P(s)/Q(s), where P and Q are polynomials, and if all the poles of Q(s) are distinct, then the inverse Laplace transform of F(s) is given by the sum of partial fraction terms.
In this case, F(s) = 1/(s(s - 3)) can be written as F(s) = A/s + B/(s - 3), where A and B are constants.
Next, we need to find the values of A and B. We can do this by manipulating the equation algebraically, getting F(s) on one side and substituting known values. In this case, we can multiply both sides by s(s-3) to get 1 = A(s-3) + Bs.
Expanding and simplifying, we get 1 = (A + B)s - 3A.
This equation holds for all s, so the coefficients of s on both sides must be equal. Equating coefficients, we get A + B = 0 and -3A = 1.
Solving these equations, we find A = -1/3 and B = 1/3.
Now, we can rewrite F(s) as F(s) = (-1/3)/s + (1/3)/(s - 3).
To find the inverse Laplace transform of F(s), we can use the table of Laplace transforms. From the table, we know that the inverse Laplace transform of 1/s is 1 and the inverse Laplace transform of 1/(s - 3) is e^3t.
Therefore, the inverse Laplace transform of F(s) = 1/(s(s - 3)) is L^(-1){F(s)} = (-1/3) + (1/3)e^(3t).