Let y=y(x) be the solution of the differential equation dy/dx + 2y = f(x), where f(x) = 1, x∈[0,1] and 0, otherwise. If y(0) = 0, then y(32) is

Solving the initial value problem, we get y = 1/2(1 - e-2x) when x∈[0,1]. We can check this by substituting this in the differential equation and checking the initial value. So, y(1) = 1/2(1 - e-2). (1) Now, for x∈(1,∞), we have e2xy = c2 (solving the differential equation separately for this interval) Using the condition found above in (1), we have c2 = e-2/2. That gives y = e-2/2e-2x for x∈(1,∞) So, for x = 32, we get y = e-2/2e-64. However, this solution seems to contain errors and doesn't match the given options. Let's reconsider the solution for x in [0,1]. The integrating factor is e∫2dx = e2x. Multiplying the differential equation by the integrating factor, we get e2x(dy/dx + 2y) = e2x. This simplifies to d/dx(ye2x) = e2x. Integrating both sides with respect to x, we have ye2x = 1/2e2x + C. Therefore, y = 1/2 + Ce-2x. Using the initial condition y(0) = 0, we get 0 = 1/2 + C, so C = -1/2. Thus, y = 1/2(1 - e-2x) for 0 ≤ x ≤ 1. For x > 1, the equation becomes dy/dx + 2y = 0, which has the solution y = Ae-2x. At x = 1, y = 1/2(1 - e-2). Therefore, A = e2/2(1 - e-2) and y = e2/2(1 - e-2)e-2x for x > 1. For x = 32, y = e2/2(1 - e-2)e-64 ≈ 0. This result does not directly match any of the options. There might be errors in the problem statement or the provided options.