By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+2 y^{\prime}+10 y=-6 e^{-t} \sin 3 t, \quad y_{0}=0, y_{0}^{\prime}=1$$

The Laplace transform is a very powerful tool when dealing with linear differential equations. It converts functions of time \( t \) into functions of a complex variable \( s \). The primary goal of the Laplace transform is to turn differential equations into algebraic equations that are easier to solve. The Laplace transform of a function \( f(t) \) is defined as: \[ \text{L} \big\{ f(t) \big\} = \bigintsss_0^\text{∞} f(t) e^{-st} dt \] Applying this transform to differential equations simplifies the process of finding solutions by converting derivatives into multiplications with \( s \). For instance, the Laplace transform of a derivative \( f'(t) \) is \[ \text{L} \big\{ f'(t) \big\} = s \text{L}\big\{ f(t) \big\} - f(0) \]

Once the differential equation is converted to an algebraic equation using the Laplace transform and solved for \( Y(s) \), we need to revert back to the time domain to find the solution \( y(t) \). This is done using the inverse Laplace transform. Essentially, the inverse Laplace transform undoes the process of the Laplace transform, changing the function of \( s \) back into a function of time. To perform an inverse Laplace transform, we often use tables of known transforms or partial fraction decomposition for more complex terms. For example, \[ \text{L}^{-1} \big\{ \frac{1}{s-a} \big\} = e^{at} \]

Differential equations involve functions and their derivatives, representing physical processes like motion or electrical circuits. Solving these equations analytically gives us insight into the behaviors of these systems. There are several methods to solve them, but using Laplace transforms is beneficial, especially for linear differential equations with constant coefficients. For example, the equation \[ y'' + 2y' + 10y = -6 e^{-t} \text{sin}(3t) \] translates into an algebraic equation in \( s \) after applying the Laplace transform, making it more tractable to solve.

Initial conditions specify the state of a system at the beginning and are crucial for finding a unique solution to differential equations. In our problem, the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \) are given. These conditions are used to determine the values of these functions when the Laplace transform is applied. By plugging these initial conditions into the transformed equation, we eliminate constants and directly solve for \( Y(s) \). This ensures our solution is specific to the given scenario.

The exponential sine function often appears in problems involving oscillatory behavior with damping, such as vibrations in mechanical systems. For example, \( e^{-t} \text{sin}(3t) \) represents a sine wave that decays over time. When applying the Laplace transform to this function, we use the formula: \[ \text{L} \big\{ e^{-at} \text{sin}(bt) \big\} = \frac{b}{(s+a)^2 + b^2} \] This helps us easily integrate it into our algebraic equation during problem-solving, simplifying our computations significantly.