Question

In each exercise, determine all equilibrium solutions (if any).\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \\\ -2\end{array}\right]\)

Short answer

Answer: The equilibrium solution for the given 2x2 linear differential equation system is y = [2, 2].

Step-by-step solution

Find a particular solution for the homogeneous equation

First, we want to find a particular solution of the given system: $$\mathbf{y}^{\prime}=\left[\begin{array}{rr}1 & -1 \\\ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \\\ -2\end{array}\right]$$ To do so, we will set \(\mathbf{y}^{\prime}\) equal to zero and solve for \(\mathbf{y}\). This will give us: $$\left[\begin{array}{rr}1 & -1 \\\ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \\\ -2\end{array}\right] = \left[\begin{array}{r}0 \\\ 0\end{array}\right]$$ Solving this equation system, we get: $$\mathbf{y} = \left[\begin{array}{r}2 \\\ 2\end{array}\right]$$

Calculate the eigenvalues and eigenvectors

Next, we need to compute the eigenvalues(\(\lambda\)) and corresponding eigenvectors of the matrix. To do this, we find the determinant of (matrix - \(\lambda I\)) and set it to 0: $$\det\left(\left[\begin{array}{rr}1-\lambda & -1 \\\ -1 & 1-\lambda\end{array}\right]\right) = 0$$ Simplifying this expression, we get: $$(1-\lambda)^{2} - (-1)(-1) = 0$$ $$\lambda^{2} - 2\lambda = 0$$ This quadratic has the solutions \(\lambda_{1} = 0\) and \(\lambda_{2} = 2\). For each eigenvalue, we can find the corresponding eigenvector by plugging the value into (matrix - \(\lambda I\))x = 0 and solving for x. For \(\lambda_{1} = 0\): $$\left[\begin{array}{rr}1 & -1 \\\ -1 & 1\end{array}\right] \mathbf{x}_1 = \left[\begin{array}{r}0 \\\ 0\end{array}\right]$$ We get the eigenvector: $$\mathbf{x}_1 = \left[\begin{array}{r}1 \\\ 1\end{array}\right]$$ For \(\lambda_{2} = 2\): $$\left[\begin{array}{rr}-1 & -1 \\\ -1 & -1\end{array}\right] \mathbf{x}_2 = \left[\begin{array}{r}0 \\\ 0\end{array}\right]$$ We get the eigenvector: $$\mathbf{x}_2 = \left[\begin{array}{r}1 \\\ -1\end{array}\right]$$

Determining equilibrium solutions

Since the eigenvalues are both real, the general solution of the system is given by: $$\mathbf{y}(t) = c_1\mathbf{x}_1 e^{\lambda_1 t} + c_2\mathbf{x}_2 e^{\lambda_2 t}$$ We can find the equilibrium solutions by setting the derivative \(y'(t)\) equal to 0 and solving for \(y(t)\). From step 1, we found the particular solution: $$\mathbf{y} = \left[\begin{array}{r}2 \\\ 2\end{array}\right]$$ which is an equilibrium solution, since the eigenvalue \(\lambda_1 = 0\). This is the only equilibrium solution since all values of \(c_1\) and \(c_2\) lead to free constants when substituting the eigenvalues. Therefore, the only equilibrium solution for this 2x2 linear differential equation system is: $$\mathbf{y} = \left[\begin{array}{r}2 \\\ 2\end{array}\right]$$

Key concepts

Differential Equations

Differential equations are mathematical tools used to model systems that change over time. These equations involve derivatives, which measure how a quantity changes with respect to another. In this exercise, we observe a first-order linear system of differential equations:
\[ \mathbf{y}^{\prime}=\left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \ -2\end{array}\right] \]
Here, \(\mathbf{y}\) is a vector that describes the state of the system, while \(\mathbf{y}'\) denotes its derivative. Models like these can represent anything from populations to electrical circuits.

In practice, understanding the solution involves finding equilibrium solutions and analyzing the system’s behavior. The equilibrium occurs when \(\mathbf{y}' = 0\), meaning the system’s state doesn’t change. We achieve this by setting up equations from the system and solving for \(\mathbf{y}\). In this example, the equilibrium solution comes out to be:
\[ \mathbf{y} = \left[\begin{array}{r}2 \ 2\end{array}\right] \]
This solution shows the point at which the system remains steady without external influence.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial for understanding linear transformations represented by matrices. In the context of differential equations, they are vital for assessing the stability of equilibrium points.

To derive eigenvalues, we transform the matrix \(A\) by subtracting \(\lambda I\) (where \(I\) is the identity matrix) and setting its determinant to zero:
\[ \det(A - \lambda I) = 0 \]
For our matrix example:
\[ \det\left(\left[\begin{array}{rr}1-\lambda & -1 \ -1 & 1-\lambda\end{array}\right]\right) = \lambda^{2} - 2\lambda = 0 \]
Solving this gives the eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 2\). Each eigenvalue corresponds to a vector (eigenvector), which aids in understanding the directions within the system that remain unchanged under transformation.

For \(\lambda_1 = 0\), plugging into the equation results in:
\[ \mathbf{x}_1 = \left[\begin{array}{r}1 \ 1\end{array}\right] \]
And for \(\lambda_2 = 2\):
\[ \mathbf{x}_2 = \left[\begin{array}{r}1 \ -1\end{array}\right] \]
These vectors provide the foundation to construct the system’s general solution, informing us about the roles these directions play in shaping the state of our system.

Homogeneous and Non-Homogeneous Systems

A linear system of differential equations can be classified as either homogeneous or non-homogeneous. This distinction lies in whether the system has a zero or non-zero constant vector on the right-hand side.

• A **homogeneous system** has the form \(\mathbf{y}^{\prime} = A\mathbf{y}\). It purely depends on the system matrix \(A\), with no external input influencing the state directly.
• A **non-homogeneous system**, like the one given in this exercise \(\mathbf{y}^{\prime}=A\mathbf{y}+\mathbf{b}\), includes an additional constant vector \(\mathbf{b}\). This term represents external influences or inputs.

To solve the non-homogeneous system, we first find a particular solution satisfying the equation with \(\mathbf{b}\). This was done by setting \(\mathbf{y}^{\prime}\) to zero and solving:
\[ \left[\begin{array}{rr}1 & -1 \ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \ -2\end{array}\right] = \left[\begin{array}{r}0 \ 0\end{array}\right] \]
The solution \(\mathbf{y} = \left[\begin{array}{r}2 \ 2\end{array}\right]\) is one such equilibrium. In conjunction with solutions from the homogeneous part, like eigenvectors, we deduce the overall behavior of the system.