Final answer:
To solve the equation x′′ + x = u(t−1) with initial conditions x(0) = 0 and x'(0) = 0, we need to consider the two cases when t < 1 and t ≥ 1 separately. For t < 1, the solution is x(t) = 0. For t ≥ 1, the solution is x(t) = t − 1.
Step-by-step explanation:
To solve the equation x′′ + x = u(t−1) with initial conditions x(0) = 0 and x'(0) = 0, we need to consider the two cases when t < 1 and t ≥ 1 separately.
For t < 1, the equation becomes x′′ + x = 0 and the general solution is x(t) = Acos(t) + Bsin(t), where A and B are constants. Since x(0) = 0, we have A = 0. Differentiating x(t), we get x'(t) = Bcos(t) - Asin(t). Since x'(0) = 0, we have B = 0. Therefore, for t < 1, x(t) = 0.
For t ≥ 1, the equation becomes x′′ + x = u(t−1), where u(t) is the unit step function. Since t ≥ 1, the unit step function u(t−1) is equal to 1. The equation simplifies to x′′ + x = 1. Using the method of undetermined coefficients, we find a particular solution x(t) = t − 1. The general solution is x(t) = Acos(t) + Bsin(t) + t − 1, where A and B are constants.
Using the initial conditions x(0) = 0 and x'(0) = 0, we can solve for the constants A and B and obtain the solution x(t) = t − 1. So, the solution to the equation x′′ + x = u(t−1) with initial conditions x(0) = 0 and x'(0) = 0 is x(t) = t − 1.