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Chapter 10: Problem 14

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{cc} 15 & -9 \\ 16 & -9 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 5 \\ 8 \end{array}\right] $$

Short Answer

Expert verified
Question: Find the solution of the initial value problem involving a first-order linear system of ordinary differential equations (ODEs) with the given matrix A and initial condition 饾懄(0) = [5, 8]. Matrix A: $$A = \begin{bmatrix} 15 & -9 \\ 16 & -9 \end{bmatrix}$$ Initial Condition: $$\mathbf{y}(0) = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$$ Solution: $$\mathbf{y}(t) = \begin{bmatrix} -21e^{12t} + 12e^{9t} \\ -7e^{12t} + 8e^{9t} \end{bmatrix}$$

Step by step solution

01

Find the eigenvalues of matrix A

First, we need to find the characteristic equation of A, which is the determinant of (A - 位I), where I is the identity matrix and 位 represents the eigenvalue. $$\text{det}(A - 位I) = \begin{vmatrix} 15 - 位 & -9 \\ 16 & -9 - 位 \end{vmatrix}$$ Now, calculate the determinant: $$\text{det}(A - 位I) = (15-位)(-9-位) - (-9)(16) = 位^2 - 6位 - 15位 + 135 + 144 = 位^2 - 21位 + 279$$
02

Solve for eigenvalues

Next, solve the characteristic equation for 位: $$位^2 - 21位 + 279 = 0$$ Notice that this quadratic equation does not factor easily. We will use the quadratic formula to solve for 位: $$位 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(279)}}{2(1)}$$ Now, calculate the discriminant inside the square root and simplify the equation to find the eigenvalues: $$\sqrt{(-21)^2 - 4(1)(279)} = 3$$ We obtain the eigenvalues as follows: $$位_1 = 12, \quad 位_2 = 9$$
03

Find the eigenvectors

Next, we need to find the eigenvectors associated with the eigenvalues 位1 and 位2. To do that, we will solve the equation (A - 位I)饾懀 = 0, where 饾懀 is the eigenvector. For 位1 = 12, $$(A - 位_1I) = \begin{bmatrix} 15 - 12 & -9 \\ 16 & -9 - 12 \end{bmatrix} = \begin{bmatrix} 3 & -9 \\ 16 & -21 \end{bmatrix}$$ Now, let's solve the system: $$\begin{cases} 3v_{11} - 9v_{12} = 0 \\ 16v_{11} - 21v_{12} = 0 \end{cases}$$ We can choose v12 = 1 arbitrary and obtain v11 = 3, so our eigenvector is: $$饾懀_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$$ For 位2 = 9, $$(A - 位_2I) = \begin{bmatrix} 15 - 9 & -9 \\ 16 & -9 - 9 \end{bmatrix} = \begin{bmatrix} 6 & -9 \\ 16 & -18 \end{bmatrix}$$ Now, let's solve the system: $$\begin{cases} 6v_{21} - 9v_{22} = 0 \\ 16v_{21} - 18v_{22} = 0 \end{cases}$$ We can choose v22 = 2 arbitrary and obtain v21 = 3, so our eigenvector is: $$饾懀_2 = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$
04

Write down the general solution

Now we have the eigenvalues and eigenvectors, we can write down the general solution for the system: $$\mathbf{y}(t) = c_1 e^{位_1 t}\mathbf{v}_1 + c_2 e^{位_2 t}\mathbf{v}_2 = c_1 e^{12t} \begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2 e^{9t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$
05

Apply the initial condition

Apply the initial condition, 饾懄(0) = [5, 8]: $$\mathbf{y}(0) = c_1 e^0 \begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2 e^0 \begin{bmatrix} 3 \\ 2 \end{bmatrix} = c_1\begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$$ Now, solve for c1 and c2: $$\begin{cases} 3c_1 + 3c_2 = 5 \\ c_1 + 2c_2 = 8 \end{cases}$$ By solving the system, we find c1 = -7 and c2 = 4.
06

Write down the solution

Finally, we can write down the solution of the initial value problem: $$\mathbf{y}(t) = -7e^{12t} \begin{bmatrix} 3 \\ 1 \end{bmatrix} + 4e^{9t} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} -21e^{12t} + 12e^{9t} \\ -7e^{12t} + 8e^{9t} \end{bmatrix}$$ That is the solution to the given 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
Eigenvalues and eigenvectors play a pivotal role in understanding and solving systems of linear differential equations. Given a square matrix, an eigenvalue is a scalar that, when subtracted from each corresponding entry along the main diagonal, turns the matrix into one having a nontrivial solution鈥攁 nonzero vector, known as an eigenvector, when multiplied by the original matrix yields the original vector scaled by the eigenvalue.

In our exercise, by working through the system \(A - \lambda I\)饾懀 = 0 where A is the matrix of coefficients, \lambda is an eigenvalue, I is the identity matrix, and v is an eigenvector, we found specific values for \lambda that are the keys to unlocking the system's behavior. The corresponding eigenvectors provide directionality to the solution space. It's fascinating how these concepts allow us to transform a dynamical system into a more manageable form by decoupling the differential equations.
Characteristic Equation
The characteristic equation is a bridge between a matrix and its eigenvalues, formulated from the determinant \( \text{det}(A - \lambda I) = 0\). Finding its roots equates to identifying the eigenvalues for the matrix A.

During our exercise, solving the characteristic equation \(\lambda^2 - 21\lambda + 279 = 0\) via the quadratic formula led to the identification of the eigenvalues \(\lambda_1 = 12\) and \(\lambda_2 = 9\). The characteristic polynomial's degree equals the matrix size and thus prescribes the number of eigenvalues and eigenvectors we can find. This algebraic approach elegantly converts a problem of dynamics into one of pure algebra.
System of Differential Equations
A system of differential equations describes a set of functions and their rates of change, concurrently. In linear algebra terms, it can be written as a single matrix differential equation \( \mathbf{y}^\prime = A\mathbf{y}\). The connectivity of the functions within the system is expressed through the matrix A, which underlines the evolution of the system over time.

Upon finding eigenvalues and eigenvectors, we craft a general solution involving exponential functions scaled by the eigenvectors. The initial conditions help determine the specific constants that make the general solution unique to the system at hand. The clear and ordered methodology displayed in solving the given initial value problem demonstrates the elegance of using these mathematical concepts to understand and describe dynamic processes.

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Most popular questions from this chapter

Describe and graph trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 9 & -3 \\ -1 & 11 \end{array}\right] \mathbf{y} $$

The matrices of the systems in exercises are singular. Describe and graph the trajectories of nonconstant solutions of the given systems. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -4 & -4 \\ 1 & 1 \end{array}\right] \mathbf{y} $$

In Exercises \(11-20\) find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system. $$ \mathbf{y}^{\prime}=\frac{1}{3}\left[\begin{array}{cc} 1 & -2 e^{-t} \\ 2 e^{t} & -1 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} e^{2 t} \\ e^{-2 t} \end{array}\right] ; \quad Y=\left[\begin{array}{cc} 2 & e^{-t} \\ e^{t} & 2 \end{array}\right] $$

Solve the initial value problem. \(\mathbf{y}^{\prime}=\frac{1}{3}\left[\begin{array}{rrr}2 & 4 & -7 \\ 1 & 5 & -5 \\ -4 & 4 & -1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}4 \\ 1 \\ 3\end{array}\right]\)

(a) Convert the scalar equation $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y=F(t) $$ into an equivalent \(n \times n\) system $$ \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t) $$ (b) Suppose (A) is normal on an interval \((a, b)\) and \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) is a fundamental set of solutions of $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y=0 $$ on \((a, b)\). Find a corresponding fundamental matrix \(Y\) for $$ \mathbf{y}^{\prime}=A(t) \mathbf{y} $$ on \((a, b)\) such that $$ y=c_{1} y_{1}+c_{2} y_{2}+\cdots+c_{n} y_{n} $$ is a solution of \((\mathrm{C})\) if and only if \(\mathrm{y}=Y \mathrm{c}\) with $$ \mathbf{c}=\left[\begin{array}{c} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \end{array}\right] $$ is a solution of (D).
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