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Chapter 10: Problem 17

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -7 & 3 \\ -3 & -1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 0 \\ 2 \end{array}\right] $$

Short Answer

Expert verified
Based on the above step by step solution, write a short answer to the question: The solution to the initial value problem for the given system of linear differential equations is: $$ \mathbf{y}(t) = -\frac{2}{3} e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \frac{4}{3} e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Step by step solution

01

Find the eigenvalues and eigenvectors of the matrix

First, let's find the eigenvalues (\(\lambda\)) of the given matrix: $$ \begin{vmatrix} A - \lambda I \end{vmatrix} = \begin{vmatrix} -7 - \lambda & 3 \\ -3 & -1 - \lambda \end{vmatrix} = \lambda^2 + 8 \lambda + 10 = (\lambda + 6)(\lambda + 2) $$ This gives us two eigenvalues: \(\lambda_1 = -6\) and \(\lambda_2 = -2\). Now we find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -6\): $$ (A - (-6)I)\mathbf{v}_1 = \begin{bmatrix} 1 & 3 \\ -3 & 5 \end{bmatrix}\mathbf{v}_1 = 0 $$ Row reducing the augmented matrix, we get: $$ \begin{bmatrix} 1 & 3 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus, \(\mathbf{v}_1 = \begin{bmatrix} 3 \\ -1 \end{bmatrix}\) is the eigenvector corresponding to the eigenvalue \(\lambda_1 = -6\). For \(\lambda_2 = -2\): $$ (A - (-2)I)\mathbf{v}_2 = \begin{bmatrix} -5 & 3 \\ -3 & -3 \end{bmatrix}\mathbf{v}_2 = 0 $$ Row reducing the augmented matrix, we get: $$ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus, \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is the eigenvector corresponding to the eigenvalue \(\lambda_2 = -2\).
02

Find the general solution

The general solution of the system of linear differential equations can be written as: $$ \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2 = C_1 e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$
03

Apply the initial conditions

Now we use the initial conditions \(\mathbf{y}(0) = \begin{bmatrix} 0 \\ 2 \end{bmatrix}\) to find the constants \(C_1\) and \(C_2\): $$ \begin{bmatrix} 0 \\ 2 \end{bmatrix} = C_1 e^{0} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 e^{0} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = C_1 \begin{bmatrix} 3 \\ -1 \end{bmatrix} + C_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ Solving the system of linear equations, we find that \(C_1 = -\frac{2}{3}\) and \(C_2 = \frac{4}{3}\).
04

Write the particular solution

Now that we have found the constants \(C_1\) and \(C_2\), we can write the particular solution of the initial value problem: $$ \mathbf{y}(t) = -\frac{2}{3} e^{-6t} \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \frac{4}{3} e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ This is the solution to the initial value problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is crucial when solving systems of linear differential equations. They reveal a lot about the nature and behavior of such systems. In a given square matrix A, eigenvalues are scalars (/lambda) that satisfy the characteristic equation det(A - (/lambda)I) = 0, where I is the identity matrix of the same dimension as A. Each (/lambda) has an associated eigenvector (/mathbf{v}), a nonzero vector that satisfies (A - (/lambda)I)(/mathbf{v}) = 0.

In our exercise, we calculate these by setting up and solving determinant equations after subtracting (/lambda) times the identity matrix from the given matrix. Row reduction helps us find a simplified system where an eigenvector can be easily identified. These eigenvectors, paired with their respective eigenvalues, provide a foundational base for constructing the general solution to differential equations.
In a teaching environment, it would be beneficial to demonstrate additional examples of finding eigenvalues and eigenvectors for a variety of matrices, with students solving these step by step to ensure strong conceptual understanding.
Systems of Linear Differential Equations
A system of linear differential equations consists of multiple equations that describe the rate of change of several interdependent variables. In matrix terms, this is often represented as (/mathbf{y}^{(/prime)}) = A(/mathbf{y}), where A is a matrix of coefficients, and (/mathbf{y}) is a vector of functions. A general solution to this system involves linear combinations of eigenvalues and corresponding eigenvectors found from the matrix A.

(/mathbf{y}(t)) = C_1 e^{(/lambda_1 t)} (/mathbf{v}_1) + C_2 e^{(/lambda_2 t)} (/mathbf{v}_2) represents the general solution where C_1 and C_2 are constants determined by the initial conditions of the problem. This form highlights how solutions to linear systems will have components that either grow or decay exponentially based on the real parts of the eigenvalues. Clear visualization through graphing software or sketches showing exponential growth or decay informed by eigenvalues can reinforce students鈥� understanding of the concepts.
Particular Solution of Differential Equations
A particular solution to a differential equation is a solution that satisfies the initial conditions of the problem. In contrast to a general solution, which includes constants representing a family of solutions, a particular solution speaks to one specific instance. To obtain it, we substitute the initial values into the general solution and solve for these constants.

(/mathbf{y}(0)) = (/mathbf{C}) provides the necessary information to determine the unique values of (/mathbf{C}). After solving for C_1 and C_2, we substitute them back into the general solution to get a formula for (/mathbf{y}(t)) that gives us the expected value of the variables for any given time t.
Students often struggle with the transition from the general to the particular solution, as this step involves solving a system of equations. Real-life contextualization of initial conditions, like populations at time zero or initial concentrations of a substance, can solidify the relevance and importance of finding particular solutions.

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Most popular questions from this chapter

Describe and graph trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 5 & -4 \\ 1 & 10 \end{array}\right] \mathbf{y} $$

Suppose the matrix $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ has a repeated eigenvalue \(\lambda_{1}\) and the associated eigenspace is one- dimensional. Let \(\mathbf{x}\) be a \(\lambda_{1}-\) eigenvector of \(A .\) Show that if \(\left(A-\lambda_{1} I\right) \mathbf{u}_{1}=\mathbf{x}\) and \(\left(A-\lambda_{1} I\right) \mathbf{u}_{2}=\mathbf{x},\) then \(\mathbf{u}_{2}-\mathbf{u}_{1}\) is parallel to \(\mathbf{x}\). Conclude from this that all vectors \(\mathbf{u}\) such that \(\left(A-\lambda_{1} I\right) \mathbf{u}=\mathbf{x}\) define the same positive and negative half-planes with respect to the line \(L\) through the origin parallel to \(\mathbf{X}\).

Plot trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -1 & 2 \end{array}\right] \mathbf{y} $$

Let \(P=P(t)\) and \(Q=Q(t)\) be the populations of two species at time \(t,\) and assume that each population would grow exponentially if the other didn't exist; that is, in the absence of competition, $$P^{\prime}=a P \quad \text { and } \quad Q^{\prime}=b Q$$ where \(a\) and \(b\) are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so (A) is replaced by $$\begin{aligned} P^{\prime} &=a P-\alpha Q \\ Q^{\prime} &=-\beta P+b Q \end{aligned}$$ where \(\alpha\) and \(\beta\) are positive constants. (Since negative population doesn't make sense, this system holds only while \(P\) and \(Q\) are both positive.) Now suppose \(P(0)=P_{0}>0\) and \(Q(0)=Q_{0}>0\) (a) For several choices of \(a, b, \alpha,\) and \(\beta,\) verify experimentally (by graphing trajectories of (A) in the \(P-Q\) plane) that there's a constant \(\rho>0\) (depending upon \(a, b, \alpha,\) and \(\beta\) ) with the following properties: (i) If \(Q_{0}>\rho P_{0},\) then \(P\) decreases monotonically to zero in finite time, during which \(Q\) remains positive. (ii) If \(Q_{0}<\rho P_{0},\) then \(Q\) decreases monotonically to zero in finite time, during which \(P\) remains positive. (b) Conclude from (a) that exactly one of the species becomes extinct in finite time if \(Q_{0} \neq \rho P_{0}\). Determine experimentally what happens if \(Q_{0}=\rho P_{0}\). (c) Confirm your experimental results and determine \(\gamma\) by expressing the eigenvalues and associated eigenvectors of $$A=\left[\begin{array}{rr} a & -\alpha \\ -\beta & b \end{array}\right]$$ in terms of \(a, b, \alpha,\) and \(\beta,\) and applying the geometric arguments developed at the end of this section.

In Exercises \(33-40\) find vectors \(\mathbf{U}\) and \(\mathbf{V}\) parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory. $$ \text { C/G } \quad \mathbf{y}^{\prime}=\left[\begin{array}{rr} -7 & 10 \\ -10 & 9 \end{array}\right] \mathbf{y} $$
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