Problem 21 - Biocalculus Calculus for the Life Sciences

Question

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{-3} & {4} \\ {-6} & {7}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{r}{-1} \\ {3}\end{array}\right]\)

Step-by-Step Solution

Answer

The solution is \( \mathbf{x}(t) = -3e^{t} \begin{bmatrix} 2 \\ 3 \end{bmatrix} + 5e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \).

Step 1: Find Eigenvalues of A

To solve the system of differential equations, first find the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) satisfy the characteristic equation \( \det(A - \lambda I) = 0 \). Compute this for \( A = \left[ \begin{array}{rr}{-3} & {4} \ {-6} & {7}\end{array}\right] \).

Step 2: Solve Characteristic Equation

The characteristic equation is: \((-3-\lambda)(7-\lambda) - (4)(-6) = 0\). Simplifying, \(\lambda^2 - 4\lambda - 3 = 0\). Solve this quadratic equation to get the eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \).

Step 3: Find Eigenvectors

For each eigenvalue, find the corresponding eigenvector. Start with \( \lambda_1 = 1 \) and solve \( (A - \lambda_1 I) \mathbf{v} = 0 \). This results in the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 2 \ 3 \end{bmatrix} \). Repeat this for \( \lambda_2 = 3 \) resulting in \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

Step 4: Form General Solution

The general solution for the system \( \frac{d \mathbf{x}}{dt} = A\mathbf{x} \) is \( \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \), where \( c_1 \) and \( c_2 \) are constants. Substituting the values gives \( \mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} 2 \ 3 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

Step 5: Apply Initial Condition

We know \( \mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} -1 \ 3 \end{bmatrix} \). Use this to solve for \( c_1 \) and \( c_2 \): \( \begin{bmatrix} -1 \ 3 \end{bmatrix} = c_1 \begin{bmatrix} 2 \ 3 \end{bmatrix} + c_2 \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Solving these equations gives \( c_1 = -3 \) and \( c_2 = 5 \).

Step 6: Write Particular Solution Using Constants

Substitute \( c_1 \) and \( c_2 \) back into the general solution to get the particular solution: \( \mathbf{x}(t) = -3 e^{t} \begin{bmatrix} 2 \ 3 \end{bmatrix} + 5 e^{3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Simplify this to obtain the final solution.

Key Concepts

Eigenvalues

Understanding eigenvalues is crucial when dealing with systems of differential equations. Eigenvalues are specific scalars associated with a square matrix that provide essential information about the system's behavior. In the context of differential equations, they help determine the stability and nature of solutions. For a given matrix \( A \), the eigenvalues \( \lambda \) are the roots of the characteristic equation \( \det(A - \lambda I) = 0 \). The determinant \( \det(A - \lambda I) \) reflects how the matrix \( A \) shifts when transformed by \( \lambda \).

• Helps assess whether solutions grow, decay, or stay constant over time
• Determines if solutions exhibit oscillatory behavior
• Critical for constructing the general solution of a differential equation system

In practice, finding eigenvalues involves solving a polynomial equation derived from \( A \). This process provides insight into the dynamic properties of a solution, such as stability and oscillation. Especially in systems where the matrix \( A \) represents a physical system, understanding the impact of each eigenvalue can guide problem-solving and system design.

Eigenvectors

Once the eigenvalues of a matrix are identified, the next step is to find its eigenvectors. These are non-zero vectors that, when multiplied by the matrix, only scale by a constant factor, which is the corresponding eigenvalue.
The relationship is given by \( (A - \lambda I) \mathbf{v} = 0 \), where \( \mathbf{v} \) is the eigenvector corresponding to eigenvalue \( \lambda \).

• Provide direction-specific scaling
• Essential for decomposing the solution space
• Form the basis of the vector space associated with a system

Finding the corresponding eigenvectors involves solving the system \( (A - \lambda I) \mathbf{v} = 0 \). This process determines the directions in which the transformation caused by \( A \) scales by \( \lambda \). For each distinct eigenvalue, there is typically a unique subspace spanned by its eigenvectors. These eigenvectors are critical to writing the general solution of differential equation systems as they define the directions in which the solution evolves.

Initial Value Problem

An initial value problem (IVP) in the context of differential equations specifies the value of the solution at a particular time, typically \( t = 0 \). This is essential for determining the unique solution of a differential equation system.
In our problem, it is given as \( \mathbf{x}(0) = \mathbf{x}_0 \).

• Ensures a unique solution conforms to given initial conditions
• Guides the computation of specific solution coefficients
• Critical in real-world applications to model state-dependent systems

To solve an IVP, once the general solution of the differential equation is found, the initial conditions are applied to determine the specific values of any constants present in the general solution. This personalization of the solution ensures that it matches the system's state at time \( t = 0 \), making it relevant for real-world scenarios like physics and engineering, where such conditions are common.

Characteristic Equation

The characteristic equation is a critical algebraic tool used to find the eigenvalues of a matrix, which further aids in solving differential equations. It is formulated by setting the determinant of the matrix \( A - \lambda I \) equal to zero.

• Facilitates identification of key properties of a matrix
• Forms the basis of determining the nature of solutions
• Provides insight into the behavior of dynamic systems

To derive the characteristic equation, subtract \( \lambda I \) from \( A \) (where \( I \) is the identity matrix) and set the determinant to zero. Solving this equation yields the eigenvalues \( \lambda \), instrumental in resolving how the system evolves over time. These eigenvalues, though derived algebraically, carry rich information about stability and the dynamics of systems characterized by the matrix \( A \). For instance, in mechanical systems, they can indicate resonant frequencies or damping characteristics.