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

When dealing with differential equations, especially those with non-homogeneous terms, the Laplace transform can be extremely useful. The general idea is to transform the differential equation into the algebraic equation. This simplifies the problem solving process significantly.
The first step is to take the Laplace transform of each term in the differential equation. For example, given the following second-order differential equation: $$ y'' + 4y' + 5y = 2e^{-2t} \u00a0cos(t), $$ we apply the Laplace transform using these properties:

• \( \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)\ \)}
• \( \mathcal{L}\{y'\} = sY(s) - y(0)\ \)}
• \( \mathcal{L}\{y\} = Y(s)\ \)}

By substituting the initial conditions \( y(0) = 0 \) and \( y'(0) = 3 \), we can reformulate the equation in terms of \(Y(s)\). This gives us an algebraic equation that is typically much easier to solve. Once we isolate \(Y(s)\), we then find its inverse Laplace transform to obtain the solution \(y(t)\). This multi-step process might seem intricate, but it efficiently simplifies solving complex computational problems.

The usefulness of Laplace transforms stems from their properties, which allow manipulation of different types of differential equations.
Some important properties include:

• **Linearity:** \( \mathcal{L}\{ af(t) + bg(t) \} = a \mathcal{L}\{ f(t) \} + b \mathcal{L}\{ g(t) \} \)}


• **Shift in Time Domain:** \( \mathcal{L}\{ e^{at} f(t) \} = F(s-a) \)}


• **Frequency Shifting:** \( \mathcal{L}{ e^{-at} f(t) \} = F(s+a) \)}


• **Differentiation in Time Domain:** \( \mathcal{L}\{ f'(t) \} = sF(s) - f(0) \) \( \mathcal{L}\{ f''(t) \} = s^2F(s) - sf(0) - f'(0) \)

Understanding these properties helps you apply the Laplace transform correctly to each term of the differential equation. For instance, consider the Laplace transform property for the exponential shift: \(\mathcal{L} \{ e^{-at} \cos(bt) \} = \frac{s + a}{(s+a)^2 + b^2} \). This allows transforming terms like \( 2e^{-2t} \cos(t) \) into a manageable algebraic form. By employing these properties correctly, you can convert a given differential equation into a simpler algebraic form that is easier to solve.
Once you understand the properties, applying them becomes straightforward, and solving differential equations using Laplace transforms becomes a systematic and streamlined process.

Once you've transformed the differential equation into an algebraic equation and solved for \(Y(s)\), the next step is to retrieve the solution in the time domain. This is done using the inverse Laplace transform.
The inverse Laplace transform converts functions from the \(s\)-domain back into the time \(t\)-domain. For instance, if we have the following equation: \( (s^2 + 4s + 5)Y(s) - 3 = 2 \cdot \frac{s+2}{(s+2)^2 + 1} \y\), we solve for \(Y(s)\), and then find the inverse Laplace transform to obtain the solution \(y(t)\).
To simplify this, we often use well-established Laplace transform tables, which provide the pairs of functions and their Laplace transforms. Knowing that:

• \( \mathcal{L} \{ e^{-at} \cos(bt) \} = \frac{s + a}{(s + a)^2 + b^2} \y\)}

matching our \( 2 \cdot \frac{s + 2}{(s + 2)^2 + 1} \), we can effectively take the inverse.
The final step involves recognizing our derived expression as a known inverse transform, thus converting back from \(Y(s)\) to \(y(t)\).
This retrieval entails combining partial fractions or using other computational methods to ensure we correctly reflect all terms in the original time domain.
Effective mastery of inverse transforms enables simple and accurate solutions, linking the original transformed problem to real-world, time-based outcomes.