Find the inverse Laplace transform of the following function:
F(s)=(9+s)/(4-s^2)
Remember your algebra? Using partial fractions, see that
F(s) = (7/4)/(s+2) - (11/4)/(s-2)
Since
L{1} = 1/s,
L{e^at f(t)} = F(s-a)
L{e^-2t} - L{e^2t} = 1/(s+2) + 1/(s-2)
I think you can take it from there.
To find the inverse Laplace transform of the function F(s) = (9 + s)/(4 - s^2), we can use the partial fraction decomposition and look up the individual inverse Laplace transforms in a standard table.
Step 1: Perform partial fraction decomposition:
F(s) = (9 + s)/(4 - s^2)
= (9 + s)/[(2 + s)(2 - s)]
The denominator can be factored into (2 + s)(2 - s). Now, we need to find the constants A and B such that:
(9 + s)/[(2 + s)(2 - s)] = A/(2 + s) + B/(2 - s)
To determine A and B, we need to find a common denominator and equate the numerators:
(9 + s) = A(2 - s) + B(2 + s)
Expanding the right-hand side and collecting like terms:
9 + s = 2A - As + 2B + Bs
Matching the coefficients of the terms with the same power of s, we get the following equations:
1. Coefficient of s^1:
1 = -A + B
2. Coefficient of s^0:
9 = 2A + 2B
Solving these equations simultaneously, we find:
A = 7/4
B = 3/4
Therefore, the partial fraction decomposition is given by:
F(s) = (7/4)/(2 + s) + (3/4)/(2 - s)
Step 2: Look up the inverse Laplace transforms:
Now, we can look up the inverse Laplace transforms of each term in the partial fraction decomposition in a standard table. The inverse Laplace transform of 1/(2 + s) is e^(-2t), and the inverse Laplace transform of 1/(2 - s) is e^(2t).
Using linearity property of Laplace transform, we have:
Inverse Laplace transform of (7/4)/(2 + s) = (7/4) * e^(-2t)
Inverse Laplace transform of (3/4)/(2 - s) = (3/4) * e^(2t)
Therefore, the inverse Laplace transform of F(s) is given by:
f(t) = (7/4) * e^(-2t) + (3/4) * e^(2t)