Answer:
Step-by-step explanation:
To find the Laplace transform of x(t) for the given differential equation:
d/dt x(t) + 2x(t) = e^(-4t) u(t) + 2u(t-1)
where u(t) represents the unit step function, we can apply the Laplace transform to both sides of the equation.
The Laplace transform of the derivative of x(t) with respect to t, denoted as L{d/dt x(t)}, can be calculated using the property of the Laplace transform:
L{d/dt x(t)} = sX(s) - x(0)
where X(s) is the Laplace transform of x(t) and x(0) is the initial value of x(t).
Using this property, the Laplace transform of the given differential equation becomes:
sX(s) - x(0) + 2X(s) = 1/(s+4) + 2e^(-s) / (s+2)
Rearranging the equation and solving for X(s), we get:
(s+2)X(s) - (x(0) - 1) = 1/(s+4) + 2e^(-s)
Now, we can apply the initial conditions to find the value of x(0). Without any specific initial conditions mentioned, we cannot determine the exact value of x(0). However, we can proceed with the general solution.
Finally, rearranging the equation and solving for X(s), we have:
X(s) = [1/(s+4) + 2e^(-s)] / (s+2) + (x(0) - 1) / (s+2)
Please note that to find the inverse Laplace transform and obtain the time-domain representation of x(t), specific initial conditions or additional information would be required.