Question

Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\to \mathrm{\infty }$. $${\mathbf{x}}^{\prime }=\left(\begin{array}{ll}1& -1\\ 5& -3\end{array}\right)\mathbf{x}$$

Short answer

Question: Express the general solution of the given system of linear differential equations using real-valued functions, given that the matrix A is: $$A=\left(\begin{array}{cc}1& -1\\ 5& -3\end{array}\right)$$ Answer: The general solution of the given system of linear differential equations can be expressed as: $$\mathbf{x}(t)=c_1{e}^{2t}\left(\begin{array}{c}1\\ 1\end{array}\right)+c_2{e}^{-4t}\left(\begin{array}{c}1\\ -5\end{array}\right)$$

Step-by-step solution

Find the Eigenvalues

First, we need to find the eigenvalues of the given matrix. To do this, we'll solve the following characteristic equation: $$\left|\begin{array}{cc}1-\lambda & -1\\ 5& -3-\lambda \end{array}\right|=0$$ Calculate the determinant of the matrix: $$(1-\lambda )((-3)-\lambda )-(-1)(5)=0$$ $${\lambda }^2+2\lambda -8=0$$ Now, solve the quadratic equation for $\lambda$: $${\lambda }_1=2,{\lambda }_2=-4$$

Find the Eigenvectors

Next, we need to find the eigenvectors associated with the eigenvalues ${\lambda }_1=2$ and ${\lambda }_2=-4$. We find the eigenvectors by solving the following system of equations: For ${\lambda }_1=2$: $$\{\begin{array}{l}(1-2)x_1-x_2=0\\ 5x_1+(-3-2)x_2=0\end{array}$$ After simplifying the equations, we can find one eigenvector to be: $${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$$ For ${\lambda }_2=-4$: $$\{\begin{array}{l}(1+4)x_1-x_2=0\\ 5x_1+(-3+4)x_2=0\end{array}$$ After simplifying the equations, we can find one eigenvector to be: $${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -5\end{array}\right)$$

Construct the General Solution

With the eigenvalues and eigenvectors at hand, we can construct the general solution of the linear system of differential equations: $$\mathbf{x}(t)=c_1{e}^{{\lambda }_{1}t}{\mathbf{v}}_1+c_2{e}^{{\lambda }_{2}t}{\mathbf{v}}_2$$ Plug in our eigenvalues and eigenvectors: $$\mathbf{x}(t)=c_1{e}^{2t}\left(\begin{array}{c}1\\ 1\end{array}\right)+c_2{e}^{-4t}\left(\begin{array}{c}1\\ -5\end{array}\right)$$ This is the general solution of the given system of linear differential equations expressed in real-valued functions.

Key concepts

Eigenvalues and Eigenvectors

In dealing with systems of differential equations, the concepts of eigenvalues and eigenvectors are invaluable. An eigenvalue is a scalar that determines the factor by which an eigenvector is scaled during a linear transformation represented by a matrix. When a matrix acts on an eigenvector, the vector is simply scaled by the eigenvalue, and its direction remains unchanged.

To find eigenvalues, we start with a square matrix from a system of differential equations and subtract the eigenvalue (represented by $\lambda$) times the identity matrix from our original matrix. When we calculate the determinant of this new matrix and set it equal to zero, we are left with the characteristic equation. Solving this equation yields the system's eigenvalues. Each eigenvalue has at least one corresponding eigenvector, which shows the direction along which the scaling occurs.

For the given exercise, calculating the determinant of the matrix minus $\lambda$ times the identity matrix and equating it to zero provided us with two eigenvalues, ${\lambda }_1=2$ and ${\lambda }_2=-4$. These are crucial for determining the system's behavior over time and are used alongside the eigenvectors to formulate the general solution.

Characteristic Equation

The characteristic equation plays a central role in understanding a system’s dynamics. It is formed by equating the determinant of the matrix, with an eigenvalue variable $\lambda$ subtracted from each entry on the main diagonal, to zero. This results in a polynomial where the degree equals the size of the square matrix.

In the exercise, the characteristic equation ${\lambda }^2+2\lambda -8=0$ is derived from the matrix by subtracting $\lambda$ from its diagonal entries and then calculating the determinant. This equation is essentially a quadratic equation in terms of $\lambda$, and solving it gives us the eigenvalues needed to proceed with finding the corresponding eigenvectors.

Understanding how to craft and solve the characteristic equation is paramount because it not only reveals the eigenvalues but also informs us about the stability and behavior of the system’s solutions over time, especially as $t$ approaches infinity. These insights are essential for predicting long-term trends in various applications such as physics and engineering systems.

General Solution of Differential Equations

The general solution of a system of linear differential equations brings together the concepts of eigenvalues and eigenvectors into a unified expression that describes the behavior of the system over time. After obtaining the eigenvalues and the corresponding eigenvectors for a matrix, we can construct the general solution by combining these elements with an exponential function of time.

In this exercise, with eigenvalues ${\lambda }_1=2$ and ${\lambda }_2=-4$, and eigenvectors ${\mathbf{v}}_1=\left(\begin{array}{c}1\text{1}\end{array}\right)$ and ${\mathbf{v}}_2=\left(\begin{array}{c}1\text{-}5\end{array}\right)$, the general solution is a linear combination of terms, each consisting of an exponential function of time multiplied by an eigenvector. The coefficients $c_1$ and $c_2$ are determined by the initial conditions of the system.

Herewith, $\mathbf{x}(t)=c_1{e}^{2t}{\mathbf{v}}_1+c_2{e}^{-4t}{\mathbf{v}}_2$, we have a powerful expression that can predict the system’s behavior for any given point in time. The positive and negative signs in the exponentials of the eigenvalues also indicate the growth or decay of the solutions respectively, which helps us visualize the system's phase portrait over time. The application of the general solution in real-world problems is widespread, ranging from natural phenomena prediction to engineering design for stable systems.