Problem 29 - Biocalculus Calculus for the Life Sciences

Question

In Exercise 10.1 .24 we considered the nongeneric system of differential equations $$\frac{d \mathbf{x}}{d t}=\left[ \begin{array}{rr}{-2} & {-1} \\ {2} & {1}\end{array}\right] \mathbf{x}$$ Theorem 2 applies to this system, and we can obtain the general solution \((8)\) in the usual way. Do so.

Step-by-Step Solution

Answer

The general solution is \( \mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} -1 \\ 3 \end{bmatrix} + c_2 e^{-4t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \).

Step 1: Write Down the System of Differential Equations

The given system of differential equations is \( \frac{d \mathbf{x}}{d t}=A\mathbf{x}\) where \( A = \left[ \begin{array}{rr}{-2} & {-1} \ {2} & {1}\end{array}\right] \) and \( \mathbf{x} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \). This compactly represents the two equations: \(\frac{dx_1}{dt} = -2x_1 - x_2 \) and \(\frac{dx_2}{dt} = 2x_1 + x_2\).

Step 2: Find the Eigenvalues of Matrix A

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). Calculate the determinant: \[ \begin{vmatrix} -2 - \lambda & -1 \ 2 & 1 - \lambda \end{vmatrix} = (-2-\lambda)(1-\lambda) - (-1)(2) = \lambda^2 + \lambda - 4 = 0. \] Factor or use the quadratic formula to find \( \lambda = 1 \) and \( \lambda = -4 \).

Step 3: Find the Eigenvectors for Each Eigenvalue

Substitute each eigenvalue back into the equation \((A - \lambda I)\mathbf{v} = 0\) to find the eigenvectors. For \( \lambda = 1 \), substitute and solve: \[ \begin{bmatrix} -3 & -1 \ 2 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}. \] This gives \( v_1 = -\frac{1}{3}v_2 \). Choose \( v_2 = 3 \), yielding the eigenvector \( \begin{bmatrix} -1 \ 3 \end{bmatrix} \). For \( \lambda = -4 \), substitute and solve: \[ \begin{bmatrix} 2 & -1 \ 2 & 5 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}. \] This gives \( v_1 = \frac{1}{2}v_2 \). Choose \( v_2 = 2 \), resulting in the eigenvector \( \begin{bmatrix} 1 \ 2 \end{bmatrix} \).

Step 4: Write the General Solution

The general solution is a linear combination of the eigenvectors multiplied by exponentials of the eigenvalues and time \(t\). This gives: \[ \mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} -1 \ 3 \end{bmatrix} + c_2 e^{-4t} \begin{bmatrix} 1 \ 2 \end{bmatrix}, \] where \(c_1\) and \(c_2\) are constants determined by initial conditions.

Key Concepts

Eigenvalues and eigenvectors

In linear algebra, eigenvalues and eigenvectors are crucial concepts, especially when solving differential equations. An eigenvalue, often denoted by \( \lambda \), is a scalar that indicates how much the direction of an eigenvector stretches or shrinks. The eigenvector is a non-zero vector that changes at most by its scalar factor \( \lambda \) when a linear transformation is applied.

In the context of solving differential equations, specifically systems like the one given in the exercise, we work with a matrix \( A \). The eigenvalues of \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). The solutions to this equation are the eigenvalues.

Once the eigenvalues are determined, each one is used to find its corresponding eigenvector \( \mathbf{v} \). This involves solving \( (A - \lambda I)\mathbf{v} = 0 \). For each eigenvalue, the associated eigenvector tells us the direction in which the transformation scales by \( \lambda \). These eigenvectors play a critical role when expressing the general solution of a differential system.

Characteristic equation

The characteristic equation is essential for finding eigenvalues. It is derived from the matrix equation \( \det(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix of the same size as \( A \) and \( \lambda \) represents the eigenvalues we aim to find.

This equation essentially determines when the matrix \( (A - \lambda I) \) becomes singular, meaning it does not have an inverse. A singular matrix leads to a system of linear equations with a non-trivial solution, which is needed to identify eigenvectors.

Applying this in practice involves:

• Setting up the matrix \( (A - \lambda I) \)
• Calculating its determinant
• Solving the resulting polynomial equation

For our exercise example, the determinant was calculated to be \( \lambda^2 + \lambda - 4 = 0 \), which when solved yielded the eigenvalues \( \lambda = 1 \) and \( \lambda = -4 \). These eigenvalues are then utilized in further calculations to determine the general solution.

General solution of differential equations

The general solution of a system of differential equations can be constructed using the eigenvalues and eigenvectors obtained from the characteristic equation. This solution gives a comprehensive expression for the variable \( \mathbf{x}(t) \) over time. The structure of this general solution involves combining the exponential functions of the eigenvalues with their respective eigenvectors.

For the exercise problem, the general solution is expressed as:

• \( c_1 e^{t} \begin{bmatrix} -1 \ 3 \end{bmatrix} \)
• \( c_2 e^{-4t} \begin{bmatrix} 1 \ 2 \end{bmatrix} \)

In essence, each part of the solution consists of an eigenvalue \( \lambda \), an exponential function \( e^{\lambda t} \) that dictates the rate of growth or decay, and an eigenvector that defines the direction of change.

The constants \( c_1 \) and \( c_2 \) are determined by initial conditions, which personalize the general solution to fit specific scenarios. This method provides a robust way of analyzing the behavior of dynamic systems over time.