Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) e^{t} $$

Short answer

Following the steps of finding eigenvalues and eigenvectors, the homogeneous solution, the particular solution, and combining them, the total general solution for the given system is: $$ \mathbf{x}(t) = C_1 e^{t}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) + \left(\begin{array}{c}{0} \\\ {2}\end{array}\right)e^{t} $$ where C_1 and C_2 are constants.

Step-by-step solution

Eigenvalues and Eigenvectors

First, we need to find the eigenvalues and eigenvectors of the matrix A: $$ A = \left(\begin{array}{ll}{2} & {-1} \\\ {3} & {-2}\end{array}\right) $$ Let's calculate the determinant of A - λI to find the eigenvalues: $$ \det(A - \lambda I) = \det \left(\begin{array}{cc}{2-\lambda} & {-1} \\\ {3} & {-2-\lambda}\end{array}\right) = (2-\lambda)(-2-\lambda) - (-1)(3) = \lambda^2 - 1 $$ Now we solve the characteristic equation for λ: $$ \lambda^2 -1 = 0 \\ \lambda = \pm 1 $$ For λ = 1, let's find the eigenvector v: $$ (A - \lambda I) v = 0 \\ \left(\begin{array}{cc}{1} & {-1} \\\ {3} & {-3}\end{array}\right)\left(\begin{array}{c}{v_1} \\\ {v_2}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ Selecting v_2 = 1, we get v_1 = 1. Therefore, the eigenvector corresponding to λ = 1 is: $$ v_1 = \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) $$ For λ = -1, let's find the eigenvector w: $$ (A - \lambda I) w = 0 \\ \left(\begin{array}{cc}{3} & {-1} \\\ {3} & {-1}\end{array}\right)\left(\begin{array}{c}{w_1} \\\ {w_2}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ Selecting w_2 = 3, we get w_1 = 1. Therefore, the eigenvector corresponding to λ = -1 is: $$ w_2 = \left(\begin{array}{c}{1} \\\ {3}\end{array}\right) $$

Homogeneous Solution

The general solution of the homogeneous system is given by: $$ \mathbf{x_h}(t) = C_1 e^{\lambda_1 t}v_1 + C_2 e^{\lambda_2 t}w_2 = C_1 e^{t}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) $$ where C_1 and C_2 are constants.

Particular Solution

To find a particular solution for the non-homogeneous term, we assume a solution of the form: $$ \mathbf{x_p}(t) = \left(\begin{array}{c}{A} \\\ {B}\end{array}\right)e^{t} $$ Thus, we have: $$ \mathbf{x_p}^{\prime}(t) = \left(\begin{array}{c}{A} \\\ {B}\end{array}\right)e^{t} $$ Now, we substitute the assumed solution and its derivative into the given system of equations and solve for A and B: $$ \left(\begin{array}{c}{A} \\\ {B}\end{array}\right)e^{t} = \left(\begin{array}{ll}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \left(\begin{array}{c}{A} \\\ {B}\end{array}\right)e^{t}+\left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) e^{t} $$ This simplifies to: $$ \left(\begin{array}{c}{A} \\\ {B}\end{array}\right) = \left(\begin{array}{ll}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \left(\begin{array}{c}{A} \\\ {B}\end{array}\right) + \left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) $$ Solving for A and B, we get: $$ \left(\begin{array}{c}{-1} \\\ {-3}\end{array}\right) = \left(\begin{array}{ll}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \left(\begin{array}{c}{A} \\\ {B}\end{array}\right) + \left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) $$ $$ \left(\begin{array}{c}{-2} \\\ {-2}\end{array}\right) = \left(\begin{array}{ll}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \left(\begin{array}{c}{A} \\\ {B}\end{array}\right) $$ Solving the system of equations, we find A = 0 and B = 2. Thus, the particular solution is: $$ \mathbf{x_p}(t) = \left(\begin{array}{c}{0} \\\ {2}\end{array}\right)e^{t} $$

Total General Solution

Now, we combine the homogeneous solution and the particular solution to get the total general solution: $$ \mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) = C_1 e^{t}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) + \left(\begin{array}{c}{0} \\\ {2}\end{array}\right)e^{t} $$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in solving systems of linear differential equations. An eigenvector represents a direction that a transformation compresses, stretches, or reverses, while its corresponding eigenvalue gives us the factor by which this transformation occurs.

To find eigenvalues, we solve the characteristic equation obtained by setting the determinant of matrix \( A - \lambda I \) to zero. For our example matrix\[ A = \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} \]finding eigenvalues involves calculating:\[ \det(A - \lambda I) = (2-\lambda)(-2-\lambda) - (-1)(3) = \lambda^2 - 1 = 0 \]This yields eigenvalues \( \lambda = \pm 1 \).Next, we find the corresponding eigenvectors by solving\( (A - \lambda I)\mathbf{v} = 0 \) for each eigenvalue:

• \( \lambda = 1 \): Results in eigenvector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
• \( \lambda = -1 \): Results in eigenvector \( \mathbf{w}_2 = \begin{pmatrix} 1 \ 3 \end{pmatrix} \).

Homogeneous and Non-Homogeneous Solutions

Differential equations can be classified based on whether they include just the homogeneous term or additional non-homogeneous terms. The general approach to solving these is by first solving the homogeneous equation and then adding to it any particular solutions to the non-homogeneous equation.

The homogeneous system associated with our given differential system is:
\( \mathbf{x}^{\prime} = A \mathbf{x} \), where \( \mathbf{x} \) is solved using the eigenvalues and eigenvectors we computed. The solution forms the basis {| \( e^{\lambda t} \)} of our general solution.

A non-homogeneous system includes an additional function. In our exercise, that function is \( \mathbf{b}(t) = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{t} \). This requires us to find a particular solution to complete the overall solution.

Matrix Algebra

Matrix algebra is fundamental when dealing with systems of linear equations, including linear differential systems. Here, matrices help represent systems compactly and enable operations like finding determinants, inverses, and eigenvalue-eigenvector calculations that simplify solutions.

In our problem, consider the matrix\[ A = \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} \]Used to define the system \( \mathbf{x}^{\prime} = A \mathbf{x} \), matrices allow a structured approach in solving for \( \mathbf{x} \), consolidating calculations in determinant form to find \( \lambda \), eigenvalues.

Solving such a matrix equation often involves setting up expressions like:\[ \det(A - \lambda I) = 0 \]which simplifies solving the system.

Particular Solution

A particular solution addresses the non-homogeneous part of the differential equation, distinguishing it by compensating for additional functions not present in earlier forms of the equation. This is necessary to fully characterize the solution set.

In our exercise, it's proposed that \( \mathbf{x_p}(t) = \begin{pmatrix} A \ B \end{pmatrix}e^{t} \). Determination of \( A \) and \( B \) involves substituting into the non-homogeneous system, yielding:\[ \begin{pmatrix} A \ B \end{pmatrix} = A \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} \begin{pmatrix} A \ B \end{pmatrix} + \begin{pmatrix} 1 \ -1 \end{pmatrix} \]Solving it, we find \( A = 0 \) and \( B = 2 \), leading to:\[ \mathbf{x_p}(t) = \begin{pmatrix} 0 \ 2 \end{pmatrix}e^{t} \]

General Solution

The general solution of a system of linear differential equations consists of both the homogeneous and particular solutions. By combining these, one can represent all possible solutions to the given differential equation system.

In our problem, the general solution is given as:\[ \mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) \]where:

• \( \mathbf{x_h}(t) = C_1 e^{t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 e^{-t} \begin{pmatrix} 1 \ 3 \end{pmatrix} \)
• \( \mathbf{x_p}(t) = \begin{pmatrix} 0 \ 2 \end{pmatrix}e^{t} \)

This accounts for every aspect of the differential equation, encompassing any solutions characterized by the constants \( C_1 \) and \( C_2 \), representing the arbitrary nature of solutions born out of the homogeneous components.