Answer:
X(s) = 2/[s³(s + 9)]
Step-by-step explanation:
Here is the complete question
Solve for the function X(s) in the Laplace domain by taking the Laplace transform of the following differential equations with given initial conditions.
dx/dt + 9x = 16t² x(0) = 0 and dx(0)/dt = 5
Solution
dx/dt + 9x = 16t²
Taking the Laplace transform of the differential equation, we have
L{dx/dt + 9x} = L{16t²}
L{dx/dt} + L{9x} = L{16t²}
L{dx/dt} + 9L{x} = 16L{t²}
sL{x} - x(0) + 9L{x} = 16L{t²}
L{x} = X(s)
L{t²} = 2!/s³
Substituting X(s) and L{t²} into the equation above, we have
sX(s) - x(0) + 9X(s) = 2!/s³
Substituting x(0) = 0 into the equation above, we have
sX(s) - x(0) + 9X(s) = 2!/s³
sX(s) - 0 + 9X(s) = 2!/s³
sX(s) + 9X(s) = 2!/s³
Factorizing X(s), we have
(s + 9)X(s) = 2!/s³
X(s) = 2/[s³(s + 9)]
So X(s) in the Laplace domain is
X(s) = 2/[s³(s + 9)]
