91Ó°ÊÓ

A system of differential equations consists of multiple equations involving derivatives of dependent variables with respect to one or more independent variables. In this exercise, the independent variable is time, denoted by \( t \), and we focus on two dependent variables \( x(t) \) and \( y(t) \).These equations describe how \( x(t) \) and \( y(t) \) change over time and how they interact with each other. The system given is:\[ \begin{aligned} \ &\frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}+\frac{d y}{d t}=0 \&\frac{d^{2} y}{d t^{2}}+\frac{d y}{d t}-4 \frac{d x}{d t}=0 \end{aligned} \]Here, these are coupled second-order linear differential equations. They need to be solved simultaneously because they depend on both \( x(t) \) and \( y(t) \) and their derivatives. Solving such systems often involves techniques like the Laplace Transform to simplify the process.

Initial conditions are specific values given at the start of the problem, which enable us to determine the unique solution of a differential equation. In the context of this system of equations, initial conditions specify the state and rate of change of the variables \( x \) and \( y \) at \( t = 0 \).For our given system:

• \( x(0) = 1 \), stating that initially, the variable \( x \) is 1.
• \( x'(0) = 0 \), indicating that \( x \) is initially not changing, as its first derivative is 0.
• \( y(0) = -1 \), specifying that \( y \) begins with the value -1.
• \( y'(0) = 5 \), showing that initially, \( y \) is increasing at the rate of 5 units per time unit.

These conditions are crucial, as they guide us to the particular solution that satisfies the problem requirements.

The inverse Laplace Transform is a mathematical technique used to convert a function in the Laplace domain back to the time domain. Once we solve the differential equations using the Laplace transform, we typically express solutions in terms of \( X(s) \) and \( Y(s) \), which are the Laplace transforms of \( x(t) \) and \( y(t) \).Performing the inverse Laplace Transform gives us back the time-domain functions \( x(t) \) and \( y(t) \). This process usually involves:

• Breaking down complex expressions into simpler fractions using partial fractions.
• Referring to known Laplace Transform pairs or tables.
• Applying linearity of the inverse operation to reconstruct the time-domain solution.

Given expressions for \( X(s) \) and \( Y(s) \), their inverse transforms return the specific solutions \( x(t) \) and \( y(t) \) needed for our system.

Solving differential equations is a process to determine functions that satisfy given equations. In our system, we use the Laplace Transform to simplify and effectively manage the equations.Here's a brief overview:- **Transform to the Laplace domain**: Replace varying equations with algebraic equations in terms of \( X(s) \) and \( Y(s) \).- **Eliminate variables**: Solve one equation for a variable. Substitute it into another to reduce the number of equations.- **Solve the system**: This might involve algebraic manipulation to isolate \( Y(s) \) or \( X(s) \).- **Back-substitute**: Use known values from one equation to find the others.Finally, employ the inverse Laplace Transform to convert and find \( x(t) \) and \( y(t) \). This turns our transformed solutions back into a comprehensible form in terms of time.