Problem 22 - Biocalculus Calculus for the Life Sciences

Question

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{0} & {1} \\ {-6} & {-5}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{-1} \\ {-2}\end{array}\right]\)

Step-by-Step Solution

Answer

The solution is \(\mathbf{x}(t) = \begin{bmatrix} e^{-2t} - 2e^{-3t} \\ -2e^{-2t} + 6e^{-3t} \end{bmatrix}\).

Step 1: Find the Eigenvalues of A

To solve the differential equation \(d \mathbf{x} / d t = A \mathbf{x}\), we first find the eigenvalues of matrix \(A\). We solve the characteristic equation \(\det(A - \lambda I) = 0\). For matrix \(A = \begin{bmatrix} 0 & 1 \ -6 & -5 \end{bmatrix}\), the characteristic equation is \(\det\left(\begin{bmatrix} -\lambda & 1 \ -6 & -5-\lambda \end{bmatrix}\right) = 0\). This simplifies to \(\lambda^2 + 5\lambda + 6 = 0\). Solving this quadratic equation gives eigenvalues \(\lambda_1 = -2\) and \(\lambda_2 = -3\).

Step 2: Find the Eigenvectors of A

Next, we find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -2\), substitute into \((A - \lambda I)\mathbf{v} = \mathbf{0}\) gives \(\begin{bmatrix} 2 & 1 \ -6 & -3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\). Solving yields the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 1 \ -2 \end{bmatrix}\). For \(\lambda_2 = -3\), solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\) gives the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ -3 \end{bmatrix}\).

Step 3: Construct the General Solution

With the eigenvalues and eigenvectors, the general solution of \(d \mathbf{x} / d t = A \mathbf{x}\) is \(\mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 1 \ -2 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 1 \ -3 \end{bmatrix}\).

Step 4: Apply Initial Condition

Use the initial condition \(\mathbf{x}(0) = \begin{bmatrix} -1 \ -2 \end{bmatrix}\) to find \(c_1\) and \(c_2\). Substituting \(t = 0\) gives \(-1 = c_1 + c_2\) and \(-2 = -2c_1 - 3c_2\). Solving these equations, we get \(c_1 = 1\) and \(c_2 = -2\).

Step 5: Write the Final Solution

Substitute \(c_1\) and \(c_2\) back into the general solution to get the specific solution: \(\mathbf{x}(t) = e^{-2t} \begin{bmatrix} 1 \ -2 \end{bmatrix} - 2 e^{-3t} \begin{bmatrix} 1 \ -3 \end{bmatrix}\). Simplifying gives \(\mathbf{x}(t) = \begin{bmatrix} e^{-2t} - 2e^{-3t} \ -2e^{-2t} + 6e^{-3t} \end{bmatrix}\).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are crucial in modeling real-world phenomena such as population growth, heat distribution, and motion of fluids. In our exercise, we are dealing with a system of first-order linear differential equations, which can be represented in matrix form as \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \). Here, \( A \) is a constant matrix, and \( \mathbf{x} \) is a vector of unknown functions depending on time \( t \).

The goal is to find the vector function \( \mathbf{x}(t) \) that satisfies both the differential equation and the initial condition, \( \mathbf{x}(0) = \mathbf{x}_0 \). Solving these types of equations usually involves finding the matrix \( A \)'s eigenvalues and eigenvectors, which simplify the system into individual differential equations that are easier to solve.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation where you are given the value of the function at a particular point, often to provide a unique solution. In our case, the initial condition \( \mathbf{x}(0) = \mathbf{x}_0 \) specifies the value of the solution at \( t = 0 \).

To solve an IVP, we first find a general solution to the differential equation. Then, we use the initial condition to find specific constants that make the solution adhere to that condition. This process ensures a unique solution tailored to the specific scenario described by the initial condition, differentiating it from other possible solutions that only satisfy the general differential equation.

Linear Algebra

Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. It is fundamental in solving differential equations like ours. The solution process involves:

• Determining the eigenvalues of the matrix \( A \) by solving the characteristic equation \( \det(A - \lambda I) = 0 \), which provides critical information on the stability and behavior of the system.
• Finding eigenvectors associated with each eigenvalue, which help in forming the general solution of the differential equation.

Matrix operations and the properties of eigenvectors and eigenvalues allow us to decouple the system of differential equations. This transformation makes it possible to solve complex systems efficiently by breaking them down into simpler, one-dimensional problems. Linear algebra techniques thus play a pivotal role in tackling differential equations and initial value problems.