A system described by the following differential equation   + 3  +2𝑦 = x(𝑡) is initially at rest. For input x(𝑡) = 2𝑢(𝑡), the output y(t) isSelect one:a. (1 + 2𝑒 −𝑡 − 2𝑒 −2𝑡 ) 𝑢(𝑡)b. (0.5 + 2𝑒 −𝑡 + 2𝑒 −2𝑡 ) 𝑢(𝑡)c. (0.5 + 𝑒 −𝑡 + 1.5𝑒 −2 ) 𝑢(𝑡)d. (1 − 2𝑒 −𝑡 + 𝑒 −2 ) 𝑢(𝑡)

To find the output y(t) for the given input x(t) = 2u(t), we need to solve the differential equation + 3y' + 2y = x(t).

1. First, let's find the homogeneous solution by setting x(t) = 0. The homogeneous equation becomes + 3y' + 2y = 0.

2. The characteristic equation is + 3r + 2 = 0. Solving this equation, we find that r = -2/3.

3. The homogeneous solution is given by y_h(t) = C1e^(-2/3t), where C1 is a constant.

4. Next, we need to find the particular solution for the given input x(t) = 2u(t). Since the input is a step function, we can assume a particular solution of the form y_p(t) = A, where A is a constant.

5. Substituting y_p(t) = A into the differential equation, we get 0 + 3(0) + 2A = 2u(t). Solving for A, we find A = 1.

6. Therefore, the particular solution is y_p(t) = 1.

7. The general solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = C1e^(-2/3t) + 1.

8. Since the system is initially at rest, we can determine the value of C1 by evaluating the initial condition y(0) = 0. Plugging in t = 0 and y(0) = 0 into the general solution, we get 0 = C1 + 1. Solving for C1, we find C1 = -1.

9. Finally, substituting C1 = -1 into the general solution, we get y(t) = -e^(-2/3t) + 1.

Therefore, the correct answer is d. (1 - 2e^(-t) + e^(-2t))u(t).