Problem 25 - Calculus for Biology and Medicine

Question

Solve the given initial-value problem. \(\left[\begin{array}{c}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{rr}4 & 7 \\ 1 & -2\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=-1\) and \(x_{2}(0)=-2\).

Step-by-Step Solution

Answer

The solution is \( x_1(t) = -\frac{35}{4} e^{5t} + \frac{3}{4} e^{-3t} \) and \( x_2(t) = -\frac{5}{4} e^{5t} - \frac{3}{4} e^{-3t} \).

Step 1: Setting Up the System of Differential Equations

We begin by setting up the system of differential equations. The given matrix equation can be written as two separate equations: \( \frac{d x_{1}}{d t} = 4 x_{1} + 7 x_{2} \) and \( \frac{d x_{2}}{d t} = x_{1} - 2 x_{2} \). These are the differential equations we need to solve.

Step 2: Solve the Characteristic Equation

To find the solution, we need the eigenvalues of the matrix. The characteristic equation is obtained by finding the determinant of \( A - \lambda I \) where \( A = \begin{pmatrix} 4 & 7 \ 1 & -2 \end{pmatrix} \) and \( I \) is the identity matrix: \[ \det \begin{pmatrix} 4-\lambda & 7 \ 1 & -2-\lambda \end{pmatrix} = (4-\lambda)(-2-\lambda) - 7 \times 1 = 0 \]. Expanding this, we get: \[ \lambda^2 - 2\lambda - 15 = 0 \].

Step 3: Find Eigenvalues

Solve the characteristic equation \( \lambda^2 - 2\lambda - 15 = 0 \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -2, c = -15 \). This gives us: \[ \lambda = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2} \] So, \( \lambda_1 = 5 \) and \( \lambda_2 = -3 \).

Step 4: Find Eigenvectors

For each eigenvalue, we need to find the corresponding eigenvector by solving \( (A - \lambda I) \mathbf{e} = 0 \). 1. For \( \lambda_1 = 5 \): \[ \begin{pmatrix} -1 & 7 \ 1 & -7 \end{pmatrix} \begin{pmatrix} e_1 \ e_2 \end{pmatrix} = 0 \]. Solving this, we find \( e_1 = 7, e_2 = 1 \), so \( \mathbf{e}_1 = \begin{pmatrix} 7 \ 1 \end{pmatrix} \). 2. For \( \lambda_2 = -3 \): \[ \begin{pmatrix} 7 & 7 \ 1 & 1 \end{pmatrix} \begin{pmatrix} e_1 \ e_2 \end{pmatrix} = 0 \]. Solving this, we find \( e_1 = -1, e_2 = 1 \), so \( \mathbf{e}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \).

Step 5: Construct the General Solution

The general solution of the system is given by: \( \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} = C_1 e^{5t} \begin{pmatrix} 7 \ 1 \end{pmatrix} + C_2 e^{-3t} \begin{pmatrix} -1 \ 1 \end{pmatrix} \) where \( C_1 \) and \( C_2 \) are constants to be determined using initial conditions.

Step 6: Apply Initial Conditions

Apply the initial conditions \( x_1(0) = -1 \) and \( x_2(0) = -2 \): \( C_1 \begin{pmatrix} 7 \ 1 \end{pmatrix} + C_2 \begin{pmatrix} -1 \ 1 \end{pmatrix} = \begin{pmatrix} -1 \ -2 \end{pmatrix} \). This results in the system: \[ 7C_1 - C_2 = -1 \] \[ C_1 + C_2 = -2 \]. Solving these equations, we find \( C_1 = -\frac{5}{4} \), \( C_2 = -\frac{3}{4} \).

Step 7: Write the Particular Solution

Substituting \( C_1 \) and \( C_2 \) back into the general solution, we get: \( \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} = -\frac{5}{4} e^{5t} \begin{pmatrix} 7 \ 1 \end{pmatrix} - \frac{3}{4} e^{-3t} \begin{pmatrix} -1 \ 1 \end{pmatrix} \). Simplifying, \( x_1(t) = -\frac{35}{4} e^{5t} + \frac{3}{4} e^{-3t} \), \( x_2(t) = -\frac{5}{4} e^{5t} - \frac{3}{4} e^{-3t} \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in describing various real-world phenomena, such as motion, heat, and fluid dynamics. In essence, these equations provide a way to model change, often in terms of rates.
Understanding differential equations involves a few key points:

• Rate of Change: The fundamental idea behind a differential equation is to express how a quantity change over time, often relating multiple variables and their rates of change.
• Systems of Differential Equations: When dealing with multiple interdependent variables, we often end up with a system of differential equations, akin to the problem given. Here, the rates of changes of two variables, \(x_1\) and \(x_2\), are described using a system represented by matrices.
• Solving Differential Equations: Methods for finding solutions vary depending on the type of differential equation. For linear systems, eigenvalues, and eigenvectors often provide a structured method to build the general solution.

In the specific problem provided, each equation describes how one variables' rate of change depends linearly on the other variables.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in the realm of linear algebra. They are particularly important in the study of systems of differential equations and play a crucial role in understanding how these systems evolve over time.
To break it down:

• Eigenvectors: An eigenvector of a matrix is a non-zero vector which only changes by a scalar factor (its eigenvalue) when that matrix is applied to it. This property is vital in simplifying complex linear transformations by restricting the focus to the most significant directions in space.
• Eigenvalues: Each eigenvector has a corresponding eigenvalue, a scalar that indicates how much the eigenvector is stretched or compressed. The pair provides insight into the dynamics of the system.

In our problem, solving the system involves finding eigenvalues that dictate how solutions behave over time. The calculated eigenvectors outline the directions along which this behavior manifests predominantly in the 2-D space. This information is crucial for expressing the general solution of the original differential equations.

Characteristic Equation

The characteristic equation is a critical tool for identifying eigenvalues of a matrix in a differential equation system. This equation arises from the determinant of the matrix \( A - \lambda I \), where \( A \) is the coefficient matrix and \( I \) is the identity matrix.
The process involves several steps:

• Formulation: Create the matrix \( A - \lambda I \), where \( \lambda \) represents potential eigenvalues.
• Determinant Calculation: Calculate the determinant of \( A - \lambda I \). Setting it to zero gives a polynomial equation.
• Solving the Equation: The solutions to this polynomial, the roots, are the eigenvalues of the matrix.

For the given problem, the characteristic equation was derived and solved, yielding eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = -3 \). These values are crucial as they provide the exponential terms in the general solution to the differential system, eventually leading to the specific behavior of the system.