Problem 43 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given system of differential equations subject to the given initial conditions. $$\begin{aligned} &\frac{d x_{1}}{d t}=-2 x_{2}, \quad \frac{d x_{2}}{d t}=2 x_{1}+4 x_{2}\\\ &x_{1}(0)=1, \quad x_{2}(0)=1 \end{aligned}$$

Step-by-Step Solution

Answer

The short answer for the given system of differential equations with initial conditions is: $$\begin{aligned} x_1(t) &= 1 + e^{-3t} \\ x_2(t) &= \frac{1}{2}e^{-3t} \end{aligned}$$

Step 1: Apply Laplace Transform to the System

Take the Laplace transform on both sides of the given differential equation system. Recall that the Laplace transform of the derivative is $$\mathcal{L}\{\frac{d x_{i}}{d t}\}(s) = sX_i(s) - x_i(0), \quad i = 1,2$$ Thus, applying the Laplace transform on both sides of the given system, we obtain: $$\begin{aligned} sX_1(s) - x_1(0) &= -2X_2(s) \\ sX_2(s) - x_2(0) &= 2X_1(s) + 4X_2(s) \end{aligned}$$ Where \(X_1(s)\) and \(X_2(s)\) are the Laplace transforms of \(x_1(t)\) and \(x_2(t)\) respectively.

Step 2: Substitute the initial conditions

Substitute the given initial conditions \(x_{1}(0)=1\) and \(x_{2}(0)=1\) into the transformed system of equations: $$\begin{aligned} sX_1(s) - 1 &= -2X_2(s) \\ sX_2(s) - 1 &= 2X_1(s) + 4X_2(s) \end{aligned}$$ Now, we have a system of linear equations involving the Laplace transforms \(X_1(s)\) and \(X_2(s)\).

Step 3: Solve the system of equations for the Laplace transforms

Use your preferred method (like substitution, elimination, or matrices) to solve the linear system for \(X_1(s)\) and \(X_2(s)\). For instance, we can isolate \(X_1(s)\) from the first equation and substitute in the second equation: $$X_1(s) = \frac{1}{s} + 2X_2(s)$$ Substitute this expression for \(X_1(s)\) into the second equation: $$sX_2(s) - 1 = 2\left(\frac{1}{s} + 2X_2(s)\right) + 4X_2(s)$$ Expanding, we obtain a linear equation in terms of \(X_2(s)\): $$X_2(s) = \frac{1}{2s+6} = \frac{1}{2(s+3)}$$ Now, substitute \(X_2(s)\) back into the equation for \(X_1(s)\): $$X_1(s) = \frac{1}{s}+2\frac{1}{2(s+3)} = \frac{1}{s} + \frac{1}{s+3}$$

Step 4: Perform inverse Laplace Transform to obtain solutions

Take the inverse Laplace transform of the results obtained in the previous step to find \(x_1(t)\) and \(x_2(t)\). Recall that the inverse Laplace transform of \(\frac{1}{s+a}\) is \(e^{-at}\): $$\begin{aligned} x_1(t) &= \mathcal{L}^{-1}\{X_1(s)\}(t) = \mathcal{L}^{-1}\left\{\frac{1}{s} + \frac{1}{s+3}\right\}(t) = e^{0t} + e^{-3t} = 1 + e^{-3t}\\ x_2(t) &= \mathcal{L}^{-1}\{X_2(s)\}(t) = \mathcal{L}^{-1}\left\{\frac{1}{2(s+3)}\right\}(t) = \frac{1}{2}e^{-3t} \end{aligned}$$ Therefore, the solutions for the given system of differential equations with initial conditions are: $$\begin{aligned} x_1(t) &= 1 + e^{-3t} \\ x_2(t) &= \frac{1}{2}e^{-3t} \end{aligned}$$

Key Concepts

System of Differential Equations

A system of differential equations involves more than one equation where the derivatives of multiple functions are related. In our exercise, we have a system with variables \( x_1(t) \) and \( x_2(t) \).
These equations describe how the functions change over time relative to one another. Our specific system is:

• \( \frac{dx_1}{dt} = -2x_2 \)
• \( \frac{dx_2}{dt} = 2x_1 + 4x_2 \)

The goal is to solve these equations simultaneously to find expressions for \( x_1(t) \) and \( x_2(t) \). They give insights into dynamic systems where variables influence each other.
Often, such systems model real-world phenomena like electrical circuits or population dynamics.
Solving these systems can be challenging, requiring techniques like Laplace Transforms which simplify handling derivatives.

Initial Conditions

Initial conditions specify the values of the functions at the start of the observation, often at \( t=0 \). They are crucial in obtaining unique solutions to differential equations.
When we solve differential equations, initial conditions help us determine specific solution paths from many possibilities. For our system:

• \( x_1(0) = 1 \)
• \( x_2(0) = 1 \)

These values influence both the character and behavior of the solutions over time. Think of initial conditions as the specific starting points on a trajectory determined by a set of rules (our differential equations). Without these, finding a precise solution is like shooting without knowing the target.

Inverse Laplace Transform

The inverse Laplace Transform allows us to move from the frequency domain back to the time domain. After solving the system of equations, we find expressions like \( X_1(s) \) and \( X_2(s) \), which are the Laplace Transforms of \( x_1(t) \) and \( x_2(t) \).
To make these results meaningful, we need to convert them back to functions of time. The inverse Laplace Transform is typically performed using tables or properties, assuming the expression is relatively simple.
For example, the inverse Laplace of \( \frac{1}{s} \) is \( 1 \), and \( \frac{1}{s+a} \) is \( e^{-at} \).
In our case:

• \( \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{1}{s+3} \right\} = 1 + e^{-3t} \)
• \( \mathcal{L}^{-1}\left\{ \frac{1}{2(s+3)} \right\} = \frac{1}{2}e^{-3t} \)

This step is crucial to answering what \( x_1(t) \) and \( x_2(t) \) are in terms of time, completing the solution of our differential equation system.

Linear System of Equations

A linear system of equations is a set of equations where each is linear in terms of the unknowns. In our problem, when applying the Laplace Transform, the original differential equations translate into linear equations involving \( X_1(s) \) and \( X_2(s) \). For our exercise:

• \( sX_1(s) - 1 = -2X_2(s) \)
• \( sX_2(s) - 1 = 2X_1(s) + 4X_2(s) \)

These equations can be solved using various methods like substitution or elimination. The solution to these linear equations gives us the Laplace Transforms of the functions, \( X_1(s) \) and \( X_2(s) \).
They represent the underlying relationships between variables in a system, analogous to equilibria in physical systems.