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 + 𝑒 −2 ) 𝑢(𝑡)  b. (0.5 + 𝑒 −𝑡 + 1.5𝑒 −2 ) 𝑢(𝑡)c. (0.5 + 2𝑒 −𝑡 + 2𝑒 −2𝑡 ) 𝑢(𝑡)d. (1 + 2𝑒 −𝑡 − 2𝑒 −2𝑡 ) 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).

Step 1: Find the homogeneous solution: The homogeneous equation is + 3y' + 2y = 0. We assume the solution to be of the form y_h(t) = e^(rt). Substituting this into the equation, we get r^2 + 3r + 2 = 0. Factoring this quadratic equation, we have (r + 1)(r + 2) = 0. So, the roots are r = -1 and r = -2.

Therefore, the homogeneous solution is y_h(t) = c1e^(-t) + c2e^(-2t), where c1 and c2 are constants.

Step 2: Find the particular solution: For the given input x(t) = 2u(t), we assume the particular solution to be of the form y_p(t) = A. Substituting this into the differential equation, we get 2 + 3(0) + 2A = 2. Solving for A, we find A = 0.

Therefore, the particular solution is y_p(t) = 0.

Step 3: Find the complete solution: The complete solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = c1e^(-t) + c2e^(-2t) + 0 = c1e^(-t) + c2e^(-2t)

Step 4: Apply initial conditions: The system is initially at rest, which means y(0) = 0. Substituting this into the complete solution, we get: 0 = c1e^(0) + c2e^(0) 0 = c1 + c2

Step 5: Determine the values of c1 and c2: Since c1 + c2 = 0, we can choose any values for c1 and c2 as long as their sum is zero. Let's choose c1 = 1 and c2 = -1.

Therefore, the complete solution becomes: y(t) = e^(-t) - e^(-2t)

Step 6: Multiply the complete solution by the input: The output y(t) is obtained by multiplying the complete solution by the input x(t) = 2u(t): y(t) = 2(e^(-t) - e^(-2t))u(t)

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