By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 t} \cos t, \quad y_{0}=0, \quad y_{0}^{\prime}=3 $$

The Laplace Transform is a powerful integral transform used to switch from the time domain to the complex frequency domain. This is especially useful in solving linear differential equations.

The Laplace transform of a function \( f(t) \) is given by the formula:
\[ \text{L}\{ f(t) \} = \int_0^{\text{\text{\hfillnity}}} e^{-st} f(t) dt \] \(L\) denotes the Laplace transform operator, and \(s\) is a complex variable.

In this exercise, the differential equation \( y'' + 4y' + 5y = 2e^{-2t} \cos(t) \) is transformed using properties of the Laplace transform:
\[ L\{y''(t)\} + 4L\{y'(t)\} + 5L\{y(t)\} = L\{2e^{-2t}\cos(t)\} \] The Laplace transform is applied to each term, providing a way to translate the problem into the frequency domain.

Initial conditions are the values of the function and its derivatives at the starting point \(t = 0\). They are essential for finding specific solutions to differential equations.

Given initial conditions for this problem are: \( y(0) = 0 \) and \( y'(0) = 3 \).

Replacing the initial conditions into the Laplace transforms of the derivatives:
\[ L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \]
\[ L\{y'(t)\} = sY(s) - y(0) \]
\[ L\{y(t)\} = Y(s) \]\( y(0) \) and \( y'(0) \) are substituted to simplify the equations:
\[ s^2Y(s) - 3 + 4sY(s) + 5Y(s) = \frac{2(s + 2)}{(s + 2)^2 + 1}\].

Initial conditions help convert the transformed differential equation into an algebraic equation.

Partial fraction decomposition simplifies complex rational expressions into simpler ones. This makes finding the inverse Laplace transform more manageable.

Given the equation:
\[ Y(s) = \frac{2(s + 2)}{((s + 2)^2 + 1)(s^2 + 4s + 5)} + \frac{3(s^2 + 4s + 5)}.\]
Decompose \(Y(s)\) into simpler terms:
For instance, \( \frac{3}{s^2 + 4s + 5} \) can be broken down:
Assume:
\[ \frac{3}{s^2 + 4s + 5} = \frac{A}{s + 2 + i} + \frac{B}{s + 2 - i}.\]

Determine the constants \(A\) and \(B\) through algebraic manipulation and substitution.
These simpler fractions can then be handled individually in the next steps.

The Inverse Laplace Transform converts expressions from the complex frequency domain back to the time domain. This step recovers the original function or solution to the differential equation.

Once \( Y(s) \) has been decomposed, apply known Laplace transform pairs.
For a term like \( \frac{3}{s^2 + 4s + 5} \), use the known inverse Laplace form:
\[ L^{-1}\{ \frac{3}{s^2 + 4s + 5} \} = 3e^{-2t} \sin(t). \]

Each term in the decomposed \( Y(s) \) undergoes this process:
- Identify the inverse pair.
- Convert each fraction back into a time-domain function.
- Combine all resulting inverse transforms.

Finally, combining all converted terms gives the complete solution for \( y(t) \).