Question

Solve the initial value problem. Eigenpairs of the coefficient matrices were determined in Exercises 1-10.\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}0 & 1 \\\ -2 & -2\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}2 \\ 2\end{array}\right]\)

Short answer

Answer: The solution of the initial value problem is $\mathbf{y}(t)=(1+i) e^{(-1+i)t} \left[\begin{array}{c} 1 \\ -1+i \end{array}\right] + (1-i) e^{(-1-i)t} \left[\begin{array}{c} 1 \\ -1-i \end{array}\right]$.

Step-by-step solution

Compute eigenvalues and eigenvectors

To compute the eigenvalues, we need to solve the characteristic equation \(|A - \lambda I| = 0\), where \(I\) is the identity matrix: $|A - \lambda I|=\left|\begin{array}{cc} - \lambda & 1 \\ -2 & -2 - \lambda \end{array}\right|=(-\lambda)(-2-\lambda) - 1(-2)=\lambda^2 + 2 \lambda +2$ We can solve the quadratic equation \(\lambda^2 + 2 \lambda +2=0\) for eigenvalues \(\lambda\): \(\Delta = 2^2 - 4(1)(2)=-4\), so the eigenvalues are complex: \(\lambda_{1,2}=\frac{-2\pm \sqrt{-4}}{2}=-1\pm i\) Now, we need to find the eigenvectors for each eigenvalue. For \(\lambda_1 = -1+i\), we can substitute it into the equation \((A - \lambda I)\vec{v} =0\): $\left[\begin{array}{cc} (-1+i) & - 1 \\ -2 & -1 -(-1+i) \end{array}\right]\left[\begin{array}{c} v_{1} \\ v_{2} \end{array}\right] =\left[\begin{array}{c} 0 \\ 0 \end{array}\right]$ We can solve this linear equation system to get the eigenvector $v_1 = \left[\begin{array}{c} 1 \\ -1+i \end{array}\right]$. Similarly, for \(\lambda_2 = -1-i\), the eigenvector is $v_2 = \left[\begin{array}{c} 1 \\ -1-i \end{array}\right]$.

Write the general solution

Using the eigenvalues and eigenvectors found in step 1, we can form the general solution \(\mathbf{y}(t)=c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2\): $\mathbf{y}(t)=c_1 e^{(-1+i)t} \left[\begin{array}{c} 1 \\ -1+i \end{array}\right] + c_2 e^{(-1-i)t} \left[\begin{array}{c} 1 \\ -1-i \end{array}\right]$

Solve for the constants

Substitute the initial value $\mathbf{y}(0)=\left[\begin{array}{c} 2 \\ 2 \end{array}\right]$ into the general solution: $\left[\begin{array}{c} 2 \\ 2 \end{array}\right] = c_1 e^{(-1+i)(0)} \left[\begin{array}{c} 1 \\ -1+i \end{array}\right] + c_2 e^{(-1-i)(0)} \left[\begin{array}{c} 1 \\ -1-i \end{array}\right]$ From this, we get the linear system of equations: \(2 = c_1 + c_2\) \(2 = (-1+i)c_1 + (-1-i)c_2\) Solving the system, we obtain \(c_1=1+i\) and \(c_2=1-i\).

Write the final solution

Substitute the constants \(c_1\) and \(c_2\) back into the general solution: $\mathbf{y}(t)=(1+i) e^{(-1+i)t} \left[\begin{array}{c} 1 \\ -1+i \end{array}\right] + (1-i) e^{(-1-i)t} \left[\begin{array}{c} 1 \\ -1-i \end{array}\right]$ This is the solution of the initial value problem.

Key concepts

Eigenvalues and Eigenvectors

In the world of linear algebra, eigenvalues and eigenvectors are incredibly powerful tools. You might encounter them when dealing with matrices, especially in areas like differential equations.
Eigenvalues, typically denoted by \( \lambda \), are scalars that, when multiplied by an eigenvector, keep the direction of the eigenvector consistent after a linear transformation represented by a matrix. Finding these values is a crucial step in solving certain types of equations.

• To find eigenvalues, solve the characteristic equation \( |A - \lambda I| = 0 \), where \( A \) is your matrix and \( I \) is the identity matrix.
• The solutions are the eigenvalues.

Once eigenvalues are calculated, the next step is to determine the eigenvectors. These vectors provide insight into how a matrix can be decomposed or factored.

• Using \( (A - \lambda I)\vec{v} = 0 \), where \( \vec{v} \) represents the eigenvector, solve the resulting system of equations for each \( \lambda \).
• Each solution gives you the direction of an eigenvector.

Eigenvalues and eigenvectors give us a way to understand complex systems in a more manageable form.

Complex Eigenvalues

In some cases, solving the characteristic equation yields complex eigenvalues. This happens when the calculated determinant results in a negative discriminant inside a square root, indicating that the roots are not real numbers.
Complex eigenvalues occur in conjugate pairs, such as \( -1+i \) and \( -1-i \). These pairs are common in certain linear differential equations, especially those involving rotation or oscillation.

• Complex conjugate pairs typically have the form \( a \pm bi \).
• The corresponding eigenvectors will have complex entries as well.

This property is useful because it relates to stability and oscillatory behavior in systems like electrical circuits or mechanical vibrations. Complex eigenvalues add a level of intricacy, turning real-world applications into fascinating contexts for analysis.
When working with these eigenvalues in differential equations, the solutions often involve expressions of the form \( e^{\text{{complex number}} \times t} \), revealing sinusoidal behaviors in systems.

Linear Differential Equations

Linear differential equations form the backbone of much of applied mathematics. They are equations that relate a function with its derivatives in a linear fashion, meaning that the highest order of the equation corresponds to the highest degree derivative.
To solve these equations, particularly when they have constant coefficients, methods involving matrices, eigenvalues, and eigenvectors come into play.

• Write your system in matrix form, identifying the coefficient matrix.
• Calculate eigenvalues and eigenvectors to aid in the construction of the general solution.

The general solution takes on the form of an exponential function, reflecting the impact of the eigenvalues on the system's behavior. Each eigenvalue contributes a distinct term to this general solution.
When it comes to initial value problems, the solutions must adhere to initial conditions which require solving for specific constants. This is done by substituting the initial values into the general solution and solving the resulting system of equations.
Understanding how these components interact provides a foundation for delving into more advanced topics in engineering, physics, and other fields utilizing mathematical analysis.