Question

In Exercises \(17-24\) solve the initial value problem. $$ \mathbf{y}^{\prime}=\frac{1}{6}\left[\begin{array}{rr} 4 & -2 \\ 5 & 2 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 1 \\ -1 \end{array}\right] $$

Short answer

Question: Find the particular solution to the given initial value problem involving a first-order linear system of differential equations. System: $$ \frac{1}{6}\begin{bmatrix} 4 & -2 \\ 5 & 2 \end{bmatrix}\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix} = \begin{bmatrix}x(t)\\y(t)\end{bmatrix},\,\,\, \begin{bmatrix}x(0)\\y(0)\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}. $$ Solution: $$ \mathbf{y}(t) = \frac{1}{2\sqrt{11}}e^{(3+\sqrt{11})t} \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix} + \frac{1}{2\sqrt{11}}e^{(3-\sqrt{11})t} \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}. $$

Step-by-step solution

Find eigenvalues and eigenvectors

We are given the matrix $$ A = \frac{1}{6}\begin{bmatrix} 4 & -2 \\ 5 & 2 \end{bmatrix}. $$ To find the eigenvalues \(\lambda\), we compute the determinant of \((A - \lambda I)\) and set it to zero: $$ \det(A - \lambda I) = \det\left(\begin{bmatrix}4-\lambda & -2\\5 & 2-\lambda\end{bmatrix}\right) = (4-\lambda)(2-\lambda) - (-2)(5) = \lambda^2 - 6\lambda + 8 - 10 = \lambda^2 - 6\lambda - 2. $$ Now, to solve the eigenvalues, we'll solve \(\lambda^2 - 6\lambda - 2 = 0\). Using the quadratic formula, $$ \lambda_{1,2} = \frac{6 \pm \sqrt{((-6)^2 - 4(-2))}}{2} = 3 \pm \sqrt{11}. $$ Now that we have the eigenvalues, we can find the corresponding eigenvectors. For \(\lambda_1 = 3 + \sqrt{11}\), plug it back into the equation \((A-\lambda_1 I)\mathbf{v}=\mathbf{0}\): $$ \begin{bmatrix} 4-(3+\sqrt{11}) & -2 \\ 5 & 2-(3+\sqrt{11}) \end{bmatrix}\begin{bmatrix}v_1 \\ v_2\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}. $$ By solving this, we get the eigenvector: \(\mathbf{v}_1 = \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix}\). For \(\lambda_2 = 3 - \sqrt{11}\), plug it back into the equation \((A-\lambda_2 I)\mathbf{v}=\mathbf{0}\): $$ \begin{bmatrix} 4-(3-\sqrt{11}) & -2 \\ 5 & 2-(3-\sqrt{11}) \end{bmatrix}\begin{bmatrix}v_1 \\ v_2\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}. $$ By solving this, we get the eigenvector: \(\mathbf{v}_2 = \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}\).

Find the general solution for the system of DE

With the eigenvalues and eigenvectors found, we can form the general solution for the system of DEs: $$ \mathbf{y}(t) = c_1 e^{(3+\sqrt{11})t} \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix} + c_2 e^{(3-\sqrt{11})t} \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}. $$

Apply the initial condition to find the particular solution

Now, we apply the initial condition \(\mathbf{y}(0) = \begin{bmatrix}1 \\ -1\end{bmatrix}\) to find the constants \(c_1\) and \(c_2\): $$ \begin{bmatrix}1 \\ -1\end{bmatrix} = c_1 e^{(3+\sqrt{11})(0)} \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix} + c_2 e^{(3-\sqrt{11})(0)} \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}. $$ This simplifies to: $$ \begin{bmatrix}1 \\ -1\end{bmatrix} = c_1 \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix} + c_2 \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}. $$ Solving this system of equations, we get \(c_1 = \frac{1}{2\sqrt{11}}\) and \(c_2 = \frac{1}{2\sqrt{11}}\). Now, plug \(c_1\) and \(c_2\) back into the general solution to get the particular solution: $$ \mathbf{y}(t) = \frac{1}{2\sqrt{11}}e^{(3+\sqrt{11})t} \begin{bmatrix}1 \\ 1 + \sqrt{11}\end{bmatrix} + \frac{1}{2\sqrt{11}}e^{(3-\sqrt{11})t} \begin{bmatrix}1 \\ 1 - \sqrt{11}\end{bmatrix}. $$ This is the solution to the initial value problem.

Key concepts

Differential Equations

Differential equations form the core of many modeling scenarios in real life, describing how quantities change over time or space. Essentially, a differential equation involves an unknown function and its derivatives. Solving these equations generally involves finding a function that satisfies the given relationship.

For example, in the exercise, the system \( \mathbf{y}^{\prime}=\frac{1}{6}\left[\begin{array}{rr} 4 & -2 \ 5 & 2 \end{array}\right] \mathbf{y} \) is a system of linear differential equations. This means that our goal is to determine how the function \( \mathbf{y}(t) \) evolves over time.

Once we solve the differential equation, it enables us to predict future behavior of the system given initial conditions. In particular, the general solution found incorporates general constants which are determined by these initial conditions.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in the study of linear transformations and systems of linear differential equations. If you've ever wondered about stability in a dynamic system or even just manipulation of matrices, these concepts come into play.

In simple terms, an eigenvalue is a scalar that transforms a non-zero vector, called an eigenvector, by multiplicative scaling during a linear transformation.

• The eigenvalue gets multiplied by the eigenvector to produce another vector that's in the same direction as the original eigenvector.
• In our solution, we found two eigenvalues: \( \lambda_1 = 3 + \sqrt{11} \) and \( \lambda_2 = 3 - \sqrt{11} \), which help us form the solution of the differential equation.
• For each eigenvalue, there exists an associated eigenvector, which in this exercise, influences the form of the general solution of our system.

Finding these eigenvalues and eigenvectors is the first step in solving systems of linear differential equations and understanding the behavior of complex systems.

Matrix Algebra

Matrix algebra serves as the mathematical foundation for solving systems of equations, including differential equations. In the context of differential equations, matrices provide a compact and efficient way to express and manipulate multiple equations simultaneously.

• One of the key matrix operations is finding the determinant, which plays a significant role in solving matrices for eigenvalues.
• Alongside, matrix addition, multiplication, and inversion are fundamental tools which allow us to transform and solve matrix equations.

Matrix algebra is woven throughout our solution as we used the matrix representation to find eigenvalues and eigenvectors by transforming our system matrix \( A \) into a simpler form. These transformations provide the stepping stones for further solutions and interpretations of any given problem.

System of Differential Equations

A system of differential equations involves two or more interrelated differential equations. These systems are essential for modeling real-world phenomena where multiple variables influence each other simultaneously.

In our exercise, we dealt with a system of two equations succinctly represented by the matrix-vector equation \( \mathbf{y}^{\prime}=\frac{1}{6}\begin{bmatrix} 4 & -2 \ 5 & 2 \end{bmatrix} \mathbf{y} \). This system captures how our 'state vector' \( \mathbf{y} \) changes over time.

• Instead of solving separate equations, we use eigenvectors and eigenvalues to find a unified solution.
• Application of initial conditions ensures that the solution curve passes through the specified point in the solution space.

In conclusion, the research of systems of differential equations is invaluable — offering insights into interactions within modeled systems whether in engineering, physics, or any other realm of applied science.