Question

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

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 𝑦(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

Find the eigenvalues of matrix A

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

Solve for eigenvalues

Next, solve the characteristic equation for λ: $$λ^2 - 21λ + 279 = 0$$ Notice that this quadratic equation does not factor easily. We will use the quadratic formula to solve for λ: $$λ = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(279)}}{2(1)}$$ 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: $$λ_1 = 12, \quad λ_2 = 9$$

Find the eigenvectors

Next, we need to find the eigenvectors associated with the eigenvalues λ1 and λ2. To do that, we will solve the equation (A - λI)𝑣 = 0, where 𝑣 is the eigenvector. For λ1 = 12, $$(A - λ_1I) = \begin{bmatrix} 15 - 12 & -9 \\ 16 & -9 - 12 \end{bmatrix} = \begin{bmatrix} 3 & -9 \\ 16 & -21 \end{bmatrix}$$ 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: $$𝑣_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$$ For λ2 = 9, $$(A - λ_2I) = \begin{bmatrix} 15 - 9 & -9 \\ 16 & -9 - 9 \end{bmatrix} = \begin{bmatrix} 6 & -9 \\ 16 & -18 \end{bmatrix}$$ 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: $$𝑣_2 = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$

Write down the general solution

Now we have the eigenvalues and eigenvectors, we can write down the general solution for the system: $$\mathbf{y}(t) = c_1 e^{λ_1 t}\mathbf{v}_1 + c_2 e^{λ_2 t}\mathbf{v}_2 = c_1 e^{12t} \begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2 e^{9t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$

Apply the initial condition

Apply the initial condition, 𝑦(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.

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.

Key concepts

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—a 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 \(A - \lambda I\)𝑣 = 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.