Question

Solve the initial value problem. Eigenpairs of the coefficient matrices were determined in Exercises 1-10.\(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}2 & 2 & 9 \\ 1 & -1 & 3 \\ -1 & -1 & -4\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}12 \\ 2 \\ 4\end{array}\right]\)

Short answer

Question: Determine the specific solution of the initial value problem for the system of linear differential equations $\mathbf{y}^{\prime} = A\mathbf{y}$, where \(A = \left[\begin{array}{rrr}2 & 2 & 9 \\1 & -1 & 3 \\-1 & -1 & -4\end{array}\right]\) and initial condition \(\mathbf{y}(0) = \left[\begin{array}{r}12 \\ 2 \\ 4\end{array}\right]\), given its eigenpairs: Eigenvalue 1: \(\lambda_1 = 1\) and eigenvector \(\mathbf{u}_1 = \left[\begin{array}{r}1 \\ 0 \\ -1\end{array}\right]\), Eigenvalue 2: \(\lambda_2 = -1\) and eigenvector \(\mathbf{u}_2 = \left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]\), Eigenvalue 3: \(\lambda_3 = -3\) and eigenvector \(\mathbf{u}_3 = \left[\begin{array}{r}3 \\ 1 \\ -2\end{array}\right]\).

Step-by-step solution

Find the general solution of the given system

First of all, let us write down the given system of linear differential equations in the matrix form:$$\mathbf{y}^{\prime} = A\mathbf{y},$$where \(A = \left[\begin{array}{rrr}2 & 2 & 9 \\1 & -1 & 3 \\-1 & -1 & -4\end{array}\right]\) and \(\mathbf{y} = \left[\begin{array}{r}y_1 \\ y_2 \\ y_3\end{array}\right]\). According to the problem, we already have the eigenpairs of matrix \(A\), which are used to find the fundamental matrix solution \(\Phi(t)\). Since the eigenpairs are known, we will have the general solution in the following form:$$\mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{u}_1 + c_2 e^{\lambda_2 t} \mathbf{u}_2 + c_3 e^{\lambda_3 t} \mathbf{u}_3,$$where \(\lambda_i\) and \(\mathbf{u}_i\) are eigenvalues and eigenvectors of matrix \(A\), and \(c_1, c_2, c_3\) are arbitrary constants.

Apply the initial condition

Now we need to apply the initial condition \(\mathbf{y}(0) = \left[\begin{array}{r}12 \\ 2 \\ 4\end{array}\right]\) to find the specific solution. By substituting \(t=0\) in the general solution we have,$$\mathbf{y}(0) = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3.$$Since the initial condition is given by \({\displaystyle \mathbf{y}(0) = \left[\begin{array}{r}12 \\ 2 \\ 4\end{array}\right]}\), we need to solve the following linear system to find the constants \(c_1, c_2, c_3\):$$c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3 = \left[\begin{array}{c}12 \\ 2 \\ 4\end{array}\right].$$After solving this linear system, we will have the specific values of \(c_1, c_2, c_3\) that satisfy the initial condition.

Write down the specific solution

Finally, we can write down the specific solution by substituting the constants \(c_1, c_2, c_3\) in the general solution given in step 1,$$\mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{u}_1 + c_2 e^{\lambda_2 t} \mathbf{u}_2 + c_3 e^{\lambda_3 t} \mathbf{u}_3.$$This is the specific solution of the initial value problem.

Key concepts

Eigenvalues and Eigenvectors

In the context of differential equations, eigenvalues and eigenvectors play a crucial role. They help us understand the behavior of systems of differential equations.

Eigenvalues are special numbers obtained from a square matrix. Mathematically, for a square matrix \(A\), an eigenvalue \(\lambda\) satisfies the equation:

\(A\mathbf{v} = \lambda \mathbf{v}\)

Here, \(\mathbf{v}\) represents an eigenvector associated with \(\lambda\). These eigenvectors are non-zero vectors that remain invariant in direction under the linear transformation represented by the matrix. Eigenvalues give insight into how these vectors grow, shrink, or rotate.

In differential equations, finding the eigenvalues and eigenvectors of the coefficient matrix helps us construct solutions. Each eigenvalue/eigenvector pair (sometimes called an eigenpair) contributes to the solution's form. They can indicate various behaviors, like stability or oscillation, within the system.

To determine these eigenpairs, one typically solves the characteristic equation, which is \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix.

Systems of Differential Equations

A system of differential equations involves multiple equations that relate the rates of change of several interdependent variables.

Consider the system given:

\(\mathbf{y}^{\prime} = A\mathbf{y}\) where \(A\) is a matrix of coefficients and \(\mathbf{y}\) defines a vector of dependent variables. Such systems are common in modeling scenarios where several quantities interact and evolve over time.

The goal here is to determine how these interacting components change with time by solving these equations. In linear systems, solutions often exploit the properties of eigenvalues and eigenvectors, as in our problem.

Matrix methods provide a structured way to find solutions. First, identify the eigenpairs of the coefficient matrix, then use these to establish a general form of the solution. This resembles deriving formulas for each component of the solution vector. Each component may be expressed as combinations of exponential functions involving the eigenvalues.

General and Specific Solutions

In differential equations, understanding both general and specific solutions is essential.

The general solution includes all possible solutions satisfying the differential equation. It often contains arbitrary constants that can take any value. For linear systems, as demonstrated in our solution, this includes terms like \(c_1 e^{\lambda_1 t} \mathbf{u}_1\), where \(c_1\) is an arbitrary constant, \(\lambda_1\) is an eigenvalue, and \(\mathbf{u}_1\) is an associated eigenvector.

The general solution forms the basis from which we derive specific solutions. A specific solution satisfies both the differential equation and a given set of initial conditions, which often relate to real-world scenarios. For example, we determine constants like \(c_1, c_2, c_3\) by applying the initial conditions \(\mathbf{y}(0)\).

This transition from general to specific solutions involves solving a system of equations derived from these initial conditions. Ultimately, this yields coefficients relevant to the situation and offers a tailored solution to the original problem.