Question

Solve the following initial value problem using Laplace transforms:

y''(t) + 4y'(t) + 4y(t) = e^(-t)

Answer 1

The initial value problem using Laplace transforms: y''(t) + 4y'(t) + 4y(t) = e^(-t), we find the solution y(t) = e^-t/4 + te^-t/2.

Step-by-step explanation:

To solve the initial value problem using Laplace transforms, we need to take the Laplace transform of both sides of the equation. Let's denote the Laplace transform of a function y(t) as Y(s). Using the properties of Laplace transforms:

s^2Y(s) + 4sY(s) + 4Y(s) = 1/(s+1)

Simplifying the equation:

Y(s) = 1/(s+1)/(s^2 + 4s + 4)

We can now factor the denominator and look up the Laplace transform pairs in a table to find the inverse Laplace transform.

The denominator factors as (s+2)^2:

Y(s) = 1/(s+1)/(s+2)^2

Using the table, we find the inverse Laplace transform:

y(t) = e^-t/4 + te^-t/2

So therefore the solution y(t) = e^-t/4 + te^-t/2.