Question

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -11 & 8 \\ -2 & -3 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 6 \\ 2 \end{array}\right] $$

Short answer

The solution to the given initial value problem is: $$ \mathbf{y}(t) = \frac{8}{3} e^{(-7 + 2\sqrt{10})t}\begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix} - \frac{4}{3} e^{(-7 - 2\sqrt{10}) t}\begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix} $$

Step-by-step solution

Compute the Eigenvalues and Eigenvectors of the Matrix

First, we need to find the eigenvalues and eigenvectors of the given matrix \(A = \begin{bmatrix} -11 & 8 \\ -2 & -3 \end{bmatrix}\). To do this, we solve the equation \(|A - \lambda I| = 0\), where \(\lambda\) are the eigenvalues and \(I\) is the identity matrix. The characteristic equation becomes: $$ \begin{vmatrix} -11-\lambda & 8 \\ -2 & -3-\lambda \end{vmatrix} = (\lambda + 11)(\lambda + 3) - 8(2) = \lambda^2 + 14\lambda + 1 = 0 $$ Solving this quadratic equation, we obtain eigenvalues \(\lambda_1 = -7 + 2\sqrt{10}\) and \(\lambda_2 = -7 - 2\sqrt{10}\). Now, we determine the corresponding eigenvectors. For \(\lambda_1\): $(A-\lambda_1 I)\mathbf{v_1} = \begin{bmatrix} -4- 2\sqrt{10}& 8 \\ -2 & 4- 2\sqrt{10} \end{bmatrix}\mathbf{v_1}=\mathbf{0}$ Choosing \(v_{12}=1\), we have \(v_{11}=\frac{1}{2+\sqrt{10}}\). For \(\lambda_2\): $(A-\lambda_2 I)\mathbf{v_2} = \begin{bmatrix} -4+ 2\sqrt{10}& 8 \\ -2 & 4+ 2\sqrt{10} \end{bmatrix}\mathbf{v_2}=\mathbf{0}$ Choosing \(v_{22}=1\), we have \(v_{21}=\frac{1}{2-\sqrt{10}}\). So we have eigenvectors \(\mathbf{v_1} = \begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix}\) and \(\mathbf{v_2} = \begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix}\) corresponding to eigenvalues \(\lambda_1\) and \(\lambda_2\) respectively.

Compute the General Solution

Now we can find the general solution for the matrix differential equation. The general solution is: \(\mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v_1} + c_2 e^{\lambda_2 t}\mathbf{v_2}\). So, $$ \mathbf{y}(t) = c_1 e^{(-7 + 2\sqrt{10})t}\begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix} + c_2 e^{(-7 - 2\sqrt{10}) t}\begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix} $$

Apply the Initial condition

Using the initial condition \(\mathbf{y}(0) = \begin{bmatrix} 6 \\ 2 \end{bmatrix}\), we can now solve for constants \(c_1\) and \(c_2\): $$ \begin{bmatrix} 6 \\ 2 \end{bmatrix} = c_1 e^{(-7 + 2\sqrt{10})(0)}\begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix} + c_2 e^{(-7 - 2\sqrt{10})(0)}\begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix} $$ Which simplifies to: $$ \begin{bmatrix} 6 \\ 2 \end{bmatrix} = c_1\begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix} + c_2\begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix} $$ Solving the system of equations, we obtain \(c_1=\frac{8}{3}\) and \(c_2=-\frac{4}{3}\).

Write the Final Solution

Now we have found the constants \(c_1\) and \(c_2\), we can write the final solution for the given initial value problem: $$ \mathbf{y}(t) = \frac{8}{3} e^{(-7 + 2\sqrt{10})t}\begin{bmatrix} \frac{1}{2+\sqrt{10}} \\ 1 \end{bmatrix} - \frac{4}{3} e^{(-7 - 2\sqrt{10}) t}\begin{bmatrix} \frac{1}{2-\sqrt{10}} \\ 1 \end{bmatrix} $$ This is the solution to the given initial value problem.

Key concepts

Eigenvalues and Eigenvectors

In matrix differential equations, understanding eigenvalues and eigenvectors plays a crucial role in solving the system. They offer essential information about the behavior of the matrix. Eigenvalues are special scalars in which there exists a vector that, when the matrix multiplies it, results in a vector that is a scaled version of the original vector. These vectors are called eigenvectors.

To calculate these, solve the equation \(|A - \lambda I| = 0\), where \(A\) is your matrix, \(I\) is the identity matrix, and \(\lambda\) represents the eigenvalues.

• For each eigenvalue \(\lambda\), substitute back into \(A - \lambda I\) to find the corresponding eigenvector.
• Check multiplicity of eigenvectors to ensure linearly independent solutions.

Determining these values reveals a lot about the matrix dynamics and how solutions will evolve over time.

Initial Value Problem

An initial value problem is a type of differential equation accompanied by conditions that specify values the solution must satisfy at a starting point, typically at time zero. These conditions are crucial because they help determine the unique solution amongst the family of potential solutions.

An initial value problem generally looks like this:

• You have a differential equation such as \( \mathbf{y}' = A\mathbf{y} \), where the given matrix \(A\) influences how \(\mathbf{y}\) changes.
• An initial condition like \( \mathbf{y}(0) = \begin{bmatrix} 6 \ 2 \end{bmatrix} \) that \(\mathbf{y}\) should satisfy at \(t=0\).

These conditions allow you to solve for constants in your general solution that specifically meet the starting criteria, ensuring the solution reflects the actual scenario.

Characteristic Equation

The characteristic equation is derived from the matrix of the system and is used to find the eigenvalues needed for the general solution. You start by setting up \(|A - \lambda I| = 0\), which is a determinant equation where \(A\) is your system's matrix, \(I\) is the identity matrix, and \(\lambda\) are the eigenvalues.

This problem gave us the characteristic equation
\[ \lambda^2 + 14\lambda + 1 = 0 \]

After solving, the roots of this equation represented the eigenvalues. These eigenvalues highlight frequencies or rates at which the functions will grow, oscillate, or decay. Hence, solving the characteristic equation is vital to understanding the system's dynamics.

General Solution

The general solution provides a comprehensive expression that describes all possible behaviors of the system under the differential equation. In matrix differential equations, the general solution often involves combining eigenvalues, eigenvectors, and exponential functions.

Given the eigenvalues \(\lambda_1\) and \(\lambda_2\) and their respective eigenvectors, the general solution looks like:
\[ \mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v_1} + c_2 e^{\lambda_2 t}\mathbf{v_2}\]

• Enter the constants \(c_1\) and \(c_2\), which are found by applying initial conditions.
• The exponential terms \(e^{\lambda t}\) direct how quickly or slowly the solution changes over time.

This formula allows you to write a particular solution that is uniquely defined by the initial conditions, giving insights into how the system evolves.