Using Laplace transforms, solve the following linear ordinary differential equation with constant coefficients:
    y''(t) + 4y'(t) + 4y(t) = e^-2t
Show your work and explain each step.

To solve the given ODE using Laplace transforms, we first need to take the Laplace transform of both sides of the equation. Recall that the Laplace transform of a function y(t) is defined as:

L{y(t)} = Y(s) = ∫₀^∞ e^(-st)y(t)dt

where s is a complex variable.

Step 1: Take the Laplace transform of both sides
The given ODE is:
y''(t) + 4y'(t) + 4y(t) = e^(-2t)

Taking the Laplace transform of both sides, we get:
L{y''(t)} + 4L{y'(t)} + 4L{y(t)} = L{e^(-2t)}

Step 2: Use properties of Laplace transforms
Now we use the following properties of Laplace transforms:

L{y'(t)} = sY(s) - y(0)
L{y''(t)} = s²Y(s) - sy(0) - y'(0)
L{e^(-at)} = 1/(s+a)

Applying these properties, we have:
[s²Y(s) - sy(0) - y'(0)] + 4[sY(s) - y(0)] + 4Y(s) = 1/(s+2)

Assume initial conditions are zero (i.e., y(0) = 0 and y'(0) = 0). This simplifies our equation to:
s²Y(s) + 4sY(s) + 4Y(s) = 1/(s+2)

Step 3: Solve for Y(s)
Factor out Y(s) from the left side of the equation:
Y(s)(s² + 4s + 4) = 1/(s+2)

Now, divide both sides by (s² + 4s + 4) to solve for Y(s):
Y(s) = 1/[(s+2)(s² + 4s + 4)]

Step 4: Partial fraction decomposition
Perform partial fraction decomposition on Y(s). We can rewrite the denominator as:
(s+2)((s+2)²)

So we have:
Y(s) = A/(s+2) + B/((s+2)²) + C/((s+2)³)

where A, B, and C are constants.

Step 5: Inverse Laplace transform
Now we need to find the inverse Laplace transform of Y(s) to get y(t). Recall that:

L⁻¹{1/(s+a)} = e^(-at)
L⁻¹{t^n*e^(-at)} = n!/(s+a)^(n+1)

Using these properties, we can find the inverse Laplace transforms of each term in Y(s):

y(t) = L⁻¹{A/(s+2) + B/((s+2)²) + C/((s+2)³)}

y(t) = A*e^(-2t) + B*t*e^(-2t) + C*(t^2)/2 * e^(-2t)

This is the solution to the given ODE. To find the specific values of A, B, and C, we would need initial conditions or boundary conditions.