Question

Solve the following ODE using the Laplace transform: y'' + 2y = u(t-1) Assuming y(0)=1 and y'(0)=2.

Answer 1

To solve the given ODE using the Laplace transform, take the Laplace transform of both sides of the equation and solve for the transformed function. Inverse Laplace transform the transformed function to obtain the solution to the original ODE.

Step-by-step explanation:

To solve the given ODE using the Laplace transform, we need to take the Laplace transform of both sides of the equation. Let's first find the Laplace transform of the differential equation y'' + 2y = u(t-1):

1. Take the Laplace transform of y''(t) using the formula L{y''(t)} = s^2Y(s) - sy(0) - y'(0), where Y(s) represents the Laplace transform of y(t).
2. Take the Laplace transform of 2y(t) using the formula L{2y(t)} = 2Y(s), where Y(s) represents the Laplace transform of y(t).
3. Take the Laplace transform of u(t-1) using the time-shifting property of the Laplace transform, which states that L{u(t-a)} = e^(-as)/s, where a represents the shift.
4. Combine the Laplace transformed terms and solve for Y(s).
5. Find the inverse Laplace transform of Y(s) to obtain the solution y(t).

By applying the Laplace transforms and solving the equation, the solution to the given ODE is y(t) = (e^(-t) - e^(-2t))u(t-1) + (e^(t) - e^(-t))u(t-3).