Find the inverse Laplace transform of the function: F(s)=1/(s(s+1)(s+2))
Use partial fractions to separate the terms. Then it should be easy.
To find the inverse Laplace transform of the function F(s) = 1/(s(s+1)(s+2)), we can use partial fraction decomposition.
Step 1: Factorize the denominator
s(s+1)(s+2) = s(s+1)(s+2)
Step 2: Write the partial fraction decomposition
1/(s(s+1)(s+2)) = A/s + B/(s+1) + C/(s+2)
Step 3: Solve for the constants A, B, and C
Multiply both sides of the equation by the denominator s(s+1)(s+2), and then plug in the corresponding values of s to solve for A, B, and C.
For s = 0, we get: 1 = A(0+1)(0+2) = 2A
A = 1/2
For s = -1, we get: 1 = B(-1)(-1+2) = B
B = 1
For s = -2, we get: 1 = C(-2)(-2+1) = -2C
C = -1/2
So, the partial fraction decomposition is:
1/(s(s+1)(s+2)) = 1/2s + 1/(s+1) - 1/2(s+2)
Step 4: Find the inverse Laplace transforms of the partial fractions
Using the table of Laplace transforms, we can find the inverse Laplace transforms of each term separately:
L{1/2s} = 1/2 (using the derivative property)
L{1/(s+1)} = e^(-t) (using the shifting property)
L{-1/2(s+2)} = -1/2 * e^(-2t) (using the shifting property)
Step 5: Combine the inverse Laplace transforms
F(t) = 1/2 - e^(-t) - 1/2 * e^(-2t)
Therefore, the inverse Laplace transform of F(s) = 1/(s(s+1)(s+2)) is F(t) = 1/2 - e^(-t) - 1/2 * e^(-2t).