Question

Use the Laplace transform to solve the given initial-value problem.

y′′+2y′+y=0, y(0)=1, y′(0)=1
y=e−t+2te−t

Answer 1

To solve the given initial-value problem using the Laplace transform, we take the Laplace transform of both sides of the differential equation, solve for Y(s), and then take the inverse Laplace transform to find y(t). The solution to the initial-value problem is y(t) = e^(-t) + 2te^(-t).

Step-by-step explanation:

To solve the given initial-value problem using the Laplace transform, we start by taking the Laplace transform of both sides of the differential equation. The Laplace transform of y''(t) is s^2Y(s) - sy(0) - y'(0), where Y(s) is the Laplace transform of y(t). Similarly, the Laplace transform of y'(t) is sY(s) - y(0). Substituting these into the differential equation and solving for Y(s), we get Y(s) = 1/(s+1)^2. Taking the inverse Laplace transform of Y(s), we find y(t) = e^(-t) + 2te^(-t). Therefore, the solution to the initial-value problem is y(t) = e^(-t) + 2te^(-t).