Use Laplace transforms to solve the following differential equation: y'' + 4y' + 5y = e^(-t), with initial conditions y(0) = 0 and y'(0) = 0. Show your work.

To solve the given differential equation using Laplace transforms, we first 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

Using the properties of Laplace transforms, we have:

L{y''(t)} = s²Y(s) - sy(0) - y'(0)
L{4y'(t)} = 4[sY(s) - y(0)]
L{5y(t)} = 5Y(s)

Now, taking the Laplace transform of both sides of the given differential equation:

s²Y(s) - sy(0) - y'(0) + 4[sY(s) - y(0)] + 5Y(s) = L{e^(-t)}

Since y(0) = 0 and y'(0) = 0, this simplifies to:

s²Y(s) + 4sY(s) + 5Y(s) = L{e^(-t)}

Now, we can factor out Y(s):

Y(s)[s² + 4s + 5] = L{e^(-t)}

To find the inverse Laplace transform, we need to get Y(s) by itself:

Y(s) = L{e^(-t)} / (s² + 4s + 5)

The denominator can be factored further:

Y(s) = L{e^(-t)} / ((s+2)² + 1)

Now, we use partial fraction decomposition to rewrite the right-hand side:

Y(s) = A / [(s+2) + i] + B / [(s+2) - i]

where A and B are complex constants.

Multiplying both sides by the denominator, we get:

L{e^(-t)} = A[(s+2) - i] + B[(s+2) + i]

Now, we can find A and B by equating coefficients of s on both sides. 

A+B=0 (coefficients of s)
-Ai+Bi=1 (constant terms)

Solving this system of equations, we get A=-B and A=i/2, so B=-i/2.

Now that we have found A and B, we can rewrite Y(s):

Y(s) = (i/2) / [(s+2) + i] - (i/2) / [(s+2) - i]

Now, we take the inverse Laplace transform of Y(s) to find y(t):

y(t) = L⁻¹{(i/2) / [(s+2) + i]} - L⁻¹{(i/2) / [(s+2) - i]}

Using the property of linearity of inverse Laplace transforms and the fact that L⁻¹{1/(s-a)} = e^(at), we get:

y(t) = (i/2)e^[-(2+i)t] - (i/2)e^[-(2-i)t]

Multiplying out the exponential terms:

y(t) = (i/2)[e^(-2t)(cos(t)+isin(t))] - (i/2)[e^(-2t)(cos(t)-isin(t))]

Combining the cosine and sine terms:

y(t) = e^(-2t)[(i/2)(cos(t)+isin(t)) - (i/2)(cos(t)-isin(t))]

Finally, simplifying the expression:

y(t) = e^(-2t)(sin(t))

So, the solution to the given differential equation with initial conditions y(0) = 0 and y'(0) = 0 is:

y(t) = e^(-2t)sin(t)