Problem 42 - Differential Equations and Linear Algebra

Question

Solve the initial-value problem \(\mathbf{x}^{\prime}=A \mathbf{x},\) where \(A=\left[\begin{array}{rr}-2 & -1 \\ 1 & -4\end{array}\right], \quad \mathbf{x}(0)=\left[\begin{array}{r}0 \\ -1\end{array}\right]\).

Step-by-Step Solution

Answer

The particular solution to the initial-value problem \(\mathbf{x}^{\prime}=A \mathbf{x}\), where \(A=\left[\begin{array}{rr}-2 & -1 \\\ 1 & -4\end{array}\right]\) and \(\mathbf{x}(0)=\left[\begin{array}{r}0 \\\ -1\end{array}\right]\) is: \(\mathbf{x}(t) = -\frac{1}{2} e^{(-3 + \sqrt{3})t} \left[\begin{array}{c}1\\1+\sqrt{3}\end{array}\right] + \frac{1}{2} e^{(-3 - \sqrt{3})t} \left[\begin{array}{c}1\\1-\sqrt{3}\end{array}\right]\)

Step 1: Calculate the eigenvalues of matrix A

First, let us find the eigenvalues of matrix A. To do that, we need to solve the characteristic equation, which is given by \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. The matrix \(A - \lambda I\) is given by: \[ A - \lambda I = \left[\begin{array}{rr} -2 - \lambda & -1 \\ 1 & -4 - \lambda \end{array}\right] \] Now, let's calculate the determinant of \(A - \lambda I\): \(\text{det}(A - \lambda I) = (-2-\lambda)(-4-\lambda) - (-1)(1)\) Expand the equation and simplify: \((\lambda^2 + 6\lambda + 7) - 1 = 0\) \(\lambda^2 + 6\lambda + 6 = 0\) Now, we have a quadratic equation to find the eigenvalues.

Step 2: Solve the quadratic equation for eigenvalues

Now, let's solve \(\lambda^2 + 6\lambda + 6 = 0\) for \(\lambda\). We can use the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where a=1, b=6, and c=6. Plugging these values into the formula: \(\lambda = \frac{-6 \pm \sqrt{6^2 - 4(1)(6)}}{2(1)}\) \(\lambda = \frac{-6 \pm \sqrt{36 - 24}}{2}\) \(\lambda = \frac{-6 \pm \sqrt{12}}{2}\) Now we have found two eigenvalues: \(\lambda_1 = -3 + \sqrt{3}\) and \(\lambda_2 = -3 - \sqrt{3}\).

Step 3: Find eigenvectors for each eigenvalue

Now, let's find the eigenvectors for each eigenvalue by solving the linear system \((A - \lambda I)v = 0\), where \(v\) is the eigenvector. For \(\lambda_1 = -3 + \sqrt{3}\): \[ \left[\begin{array}{cc} 1-\sqrt{3} & -1 \\ 1 & -1-\sqrt{3} \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \] We can find one eigenvector for \(\lambda_1\): \(v_1 = \left[\begin{array}{c}1\\1 + \sqrt{3}\end{array}\right]\) For \(\lambda_2 = -3 - \sqrt{3}\): \[ \left[\begin{array}{cc}1+\sqrt{3} & -1 \\ 1 & -1+\sqrt{3} \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \] We can find one eigenvector for \(\lambda_2\): \(v_2 = \left[\begin{array}{c}1\\1 - \sqrt{3}\end{array}\right]\)

Step 4: Write the general solution

Now we have eigenvalues and eigenvectors to form the general solution for the system: \(\mathbf{x}(t) = c_1 e^{(-3 + \sqrt{3})t} \left[\begin{array}{c}1\\1+\sqrt{3}\end{array}\right] + c_2 e^{(-3 - \sqrt{3})t} \left[\begin{array}{c}1\\1-\sqrt{3}\end{array}\right]\)

Step 5: Use the initial condition to find the particular solution

Applying the initial condition \(\mathbf{x}(0)=\left[\begin{array}{r}0 \\ -1\end{array}\right]\), we have: \(c_1\left[\begin{array}{c}1\\1+\sqrt{3}\end{array}\right] + c_2\left[\begin{array}{c}1\\1-\sqrt{3}\end{array}\right] = \left[\begin{array}{r}0 \\ -1\vphantom{\sqrt{3}}\end{array}\right]\) Solving this linear system, we obtain \(c_1 = -\frac{1}{2}\) and \(c_2 = \frac{1}{2}\).

Step 6: Write the particular solution

Finally, we have the particular solution \(\mathbf{x}(t)\): \(\mathbf{x}(t) = -\frac{1}{2} e^{(-3 + \sqrt{3})t} \left[\begin{array}{c}1\\1+\sqrt{3}\end{array}\right] + \frac{1}{2} e^{(-3 - \sqrt{3})t} \left[\begin{array}{c}1\\1-\sqrt{3}\end{array}\right]\) This is the solution to the initial-value problem.

Key Concepts

Initial Value Problem

An initial value problem is a differential equation that requires solving over a given interval with known starting values. In this context, we have a system of equations represented by

• a differential equation: \[ \mathbf{x}^{\prime} = A \mathbf{x} \]
• and an initial condition: \[ \mathbf{x}(0)=\left[\begin{array}{r}0 \ -1\end{array}\right] \]

Effectively, you need to find the function \( \mathbf{x}(t) \) which satisfies the differential equation with \( \mathbf{x}(0) \) given. This involves determining the behavior of \( \mathbf{x}(t) \) for \( t \geq 0 \) based on the matrix \( A \) and the initial vector.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental in solving systems of differential equations like the one given in the problem. To find them, we first calculate the eigenvalues of the matrix \( A \) by solving the characteristic equation

• \[ \text{det}(A - \lambda I) = 0 \]
• For our specific case: \[ A - \lambda I = \left[\begin{array}{rr} -2 - \lambda & -1 \ 1 & -4 - \lambda \end{array}\right] \]

Solving this gives us the eigenvalues \( \lambda_1 = -3 + \sqrt{3} \) and \( \lambda_2 = -3 - \sqrt{3} \). Each eigenvalue has a corresponding eigenvector, which we find by solving

• \[ (A - \lambda I)v = 0 \]

For example, an eigenvector corresponding to \( \lambda_1 \) is \[\left[\begin{array}{c}1\1 + \sqrt{3}\end{array}\right]\] These provide the foundational blocks to express \( \mathbf{x}(t) \).

Matrix Solutions

To solve such an initial value problem, we first express the solution in terms of the exponential of eigenvalues. The general solution form becomes:

• \[ \mathbf{x}(t) = c_1e^{\lambda_1 t}v_1 + c_2e^{\lambda_2 t}v_2 \]

- This combines the eigenvectors \( v_1 \) and \( v_2 \) with their respective exponential expressions. - By adjusting constants \( c_1 \) and \( c_2 \) using initial conditions, we determine a particular solution that fits the specific starting values.

• For \( \mathbf{x}(0) = \left[\begin{array}{r}0 \ -1\end{array}\right] \)

, solving this yields specific values for \( c_1 \) and \( c_2 \), completing the particular solution.

Linear Systems

In mathematics, linear systems represent equations involving linear expressions. Here, we deal with a linear system in the context of matrices and vectors. Act as a backbone for representing relationships in

• physics
• engineering
• economics

Using Matrices in Linear Systems

- Matrix \( A \) determines the relationship between the current state \( \mathbf{x}(t) \) and its rate of change \( \mathbf{x}^{\prime}(t) \). - The simplicity lies in the linear nature of these systems, despite their deep application potential. Using matrices allows us to handle systems efficiently and uniformly, which is especially beneficial in computational solutions.