Question

Find the particular solution of the linear system that satisfies the stated initial conditions. $$ \begin{aligned} &x^{\prime}=x-2 y \\ &y^{\prime}=8 x-7 y \\ &x(0)=6, y(0)=8 \end{aligned} $$

Short answer

The particular solution of the linear system that satisfies the initial conditions is: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = 3 e^{-3t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 3 e^{5t} \begin{pmatrix} -1 \\ 2 \end{pmatrix} \]

Step-by-step solution

Write down the matrix form of the linear system

The first step is to represent the given system of linear differential equations in matrix form. The system can be written as: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 8 & -7 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \]

Find the eigenvalues and eigenvectors of the matrix

Next, we'll find the eigenvalues and eigenvectors of the matrix. The characteristic equation is given by: \[ \det(A - \lambda I) = \det\begin{pmatrix} 1-\lambda & -2 \\ 8 & -7-\lambda \end{pmatrix}=(1-\lambda)(-7-\lambda)-(-2)(8)=\lambda^2+6\lambda-15 \] Solving for the eigenvalues, we get: \[ \lambda^2+6\lambda-15=(\lambda+3)(\lambda-5)=0, \] So the eigenvalues are \(\lambda_1 = -3\) and \(\lambda_2 = 5\). Now we need to find the eigenvectors associated with these eigenvalues. For \(\lambda_1 = -3\), we have: \[ A - \lambda_1 I = \begin{pmatrix} 4 & -2 \\ 8 & -4 \end{pmatrix} \] The eigenvector for \(\lambda_1 = -3\) can be found by solving the following linear system: \[ \begin{pmatrix} 4 & -2 \\ 8 & -4 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can simplify this system and find the eigenvector: \[ c_1 = \frac{1}{2} c_2 \] For example, we can choose \(c_2 = 2\), so the eigenvector is \(v_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\). For \(\lambda_2 = 5\), we have: \[ A - \lambda_2 I = \begin{pmatrix} -4 & -2 \\ 8 & -12 \end{pmatrix} \] The eigenvector for \(\lambda_2 = 5\) can also be found by solving the following linear system: \[ \begin{pmatrix} -4 & -2 \\ 8 & -12 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can simplify this system and find the eigenvector: \[ c_1 = -\frac{1}{2} c_2 \] For example, we can choose \(c_2 = 2\), so the eigenvector is \(v_2 = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\).

Write the general solution and apply the initial conditions

Now that we have the eigenvalues and eigenvectors, we can write the general solution of the given system of linear differential equations in the form: \[ \begin{pmatrix} x \\ y \end{pmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2 = c_1 e^{-3t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{5t} \begin{pmatrix} -1 \\ 2 \end{pmatrix} \] We are given the initial conditions \(x(0) = 6\) and \(y(0) = 8\). We can use these initial conditions to solve for the constants \(c_1\) and \(c_2\): \[ \begin{pmatrix} 6 \\ 8 \end{pmatrix} = c_1 e^{0} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{0} \begin{pmatrix} -1 \\ 2 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 \begin{pmatrix} -1 \\ 2 \end{pmatrix} \] By solving this linear system, we find \(c_1 = 3\) and \(c_2 = 3\).

Write the particular solution

Finally, we can write the particular solution of the given linear system using the calculated constants: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = 3 e^{-3t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 3 e^{5t} \begin{pmatrix} -1 \\ 2 \end{pmatrix} \] This is the particular solution that satisfies the given initial conditions.

Key concepts

Initial Conditions

In the study of differential equations, **initial conditions** are crucial, as they allow us to find a specific solution to a differential equation from a set of many possible solutions. Initial conditions specify the values of the solution at a certain point, usually at the start.
For this exercise, the initial conditions given were \( x(0) = 6 \) and \( y(0) = 8 \). These values act like a starting point that anchor the system of equations to a particular solution.
Understanding initial conditions is important because:

• They adjust the general solution into a specific solution that meets the given conditions.
• They help in solving the constants involved in the solution, making it possible to describe the behavior of the system precisely at any point in time.

This is essential for applying the mathematical model to real-world situations.

Matrix Representation

Transforming a system of linear differential equations into a **matrix form** is a powerful tool because it simplifies computations and allows the use of matrix algebra techniques. In the given system:\[x' = x - 2y \y' = 8x - 7y\]we represent it in matrix form as:\[\begin{pmatrix}x' \y'\end{pmatrix} =\begin{pmatrix}1 & -2 \8 & -7\end{pmatrix}\begin{pmatrix}x \y\end{pmatrix}\]In this representation:

• The matrix contains the coefficients of \(x\) and \(y\) from the equations.
• The variables and their derivatives are organized in column vectors.

These matrices help in visualizing systems and using matrix operations, like multiplication and finding determinants, simplifies further calculations in the process of solving the system.

Eigenvalues and Eigenvectors

**Eigenvalues and eigenvectors** are pivotal in understanding the dynamics of systems. They provide insight into system behavior such as stability and response over time. In our case, these are derived from the matrix representation of the system.
We find the characteristic polynomial \( \det(A - \lambda I) \) of the matrix and solve it as:\[\begin{vmatrix} 1-\lambda & -2 \8 & -7-\lambda \end{vmatrix} = \lambda^2 + 6\lambda - 15 = 0\]This gives us **eigenvalues** \(\lambda_1 = -3\) and \(\lambda_2 = 5\). For each eigenvalue, we find the associated eigenvector by solving the system \((A - \lambda I)v = 0\). This involves linear algebra techniques like Gaussian elimination or substitution.

• **Eigenvalue \(\lambda_1 = -3\)** leads to an eigenvector \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \).
• **Eigenvalue \(\lambda_2 = 5\)** leads to an eigenvector \( \begin{pmatrix} -1 \ 2 \end{pmatrix} \).

These elements are essential for constructing the general solution of differential equations.

General Solution

The **general solution** of a system of differential equations encompasses all possible solutions and provides a framework to derive particular solutions under specific conditions. For systems like ours, this solution is a combination of the modes found by eigenvalue decomposition.
The general solution for the matrix-based system is expressed using the eigenvalues and eigenvectors:\[\begin{pmatrix}x(t) \y(t)\end{pmatrix} = c_1 e^{\lambda_1 t} \begin{pmatrix} 1 \2 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} -1 \2 \end{pmatrix}\]where **\(e^{\lambda t}\)** introduces the exponential growth or decay dictated by the eigenvalues, and **\(c_1\)** and **\(c_2\)** are constants determined using initial conditions.

• The general form captures the complete dynamical behavior of the system.
• We use the initial conditions to solve for \(c_1\) and \(c_2\) to find the particular solution: \(c_1 = 3\) and \(c_2 = 3\) in the given exercise.

Thus, the system's particular solution is derived and matches the initial conditions perfectly, illustrating the process from a systematic and mathematical standpoint.