Question

In Exercises \(78-88\) find a fundamental set of solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\). Solve the initial value problem with \(\mathbf{x}(0)=\mathbf{x}_{0}\). $$ A=\left[\begin{array}{rrr} -3 & 1 & 0 \\ -5 & 1 & 0 \\ -9 & 3 & -2 \end{array}\right], \quad \mathbf{x}_{0}=\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right] $$

Short answer

The solution is \( \mathbf{x}(t) = e^{-2t} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -2e^{-2t} - 2t e^{-2t} \end{bmatrix} \).

Step-by-step solution

Find the Eigenvalues

To find the fundamental set of solutions, we begin by finding the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). For matrix \( A \), this becomes: \[\det\left(\begin{array}{rrr} -3 - \lambda & 1 & 0 \ -5 & 1 - \lambda & 0 \ -9 & 3 & -2 - \lambda \end{array}\right) = 0\] Expanding the determinant and solving the resulting polynomial equation gives \( \lambda_1 = -1 \) and a repeated root \( \lambda_2 = \lambda_3 = -2 \).

Find the Eigenvectors

For each eigenvalue found, we find the corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). Start with \( \lambda_1 = -1 \):\[\left(\begin{array}{rrr}-2 & 1 & 0 \-5 & 2 & 0 \-9 & 3 & -1\end{array}\right)\left(\begin{array}{c} x_1 \ x_2 \ x_3 \end{array}\right)= \left(\begin{array}{c} 0 \ 0 \ 0 \end{array}\right)\]Solving this gives the eigenvector \( \mathbf{v}_1 = \left(\begin{array}{c} 1 \ 2 \ 0 \end{array}\right) \).For the repeated eigenvalue \( \lambda_2 = \lambda_3 = -2 \), solve the system:\[\left(\begin{array}{rrr}-1 & 1 & 0 \-5 & 3 & 0 \-9 & 3 & 0\end{array}\right)\left(\begin{array}{c} x_1 \ x_2 \ x_3 \end{array}\right)= \left(\begin{array}{c} 0 \ 0 \ 0 \end{array}\right)\]This results in the eigenvector \( \mathbf{v}_2 = \left(\begin{array}{c} 1 \ 1 \ 0 \end{array}\right) \). Further vector \( \mathbf{v}_3 \) due to the Jordan form gives \( \left(\begin{array}{c} 0 \ 0 \ 1 \end{array}\right) \).

Form the General Solution

Using the eigenvalues and corresponding eigenvectors, the general solution to the differential equation \( \mathbf{x}' = A \mathbf{x} \) is constructed as:\[\mathbf{x}(t) = c_1 e^{-t} \begin{bmatrix} 1 \ 2 \ 0 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + c_3 \begin{bmatrix} t e^{-2t} \ t e^{-2t} \ e^{-2t} \end{bmatrix}\]

Apply Initial Conditions

Insert the initial condition \( \mathbf{x}(0) = \begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} \) to solve for the constants \( c_1, c_2, \) and \( c_3 \):\[\begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} = c_1 \begin{bmatrix} 1 \ 2 \ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}\]Solving this system gives \( c_1 = 0 \), \( c_2 = 1 \), and \( c_3 = -2 \).

Write the Solution to the IVP

With \( c_1, c_2, \) and \( c_3 \) found, write the solution to the initial value problem:\[\mathbf{x}(t) = e^{-2t} \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + \begin{bmatrix} 0 \ 0 \ -2e^{-2t} \end{bmatrix} + \begin{bmatrix} 0 \ 0 \ -2t e^{-2t} \end{bmatrix}\]

Key concepts

Eigenvalues

Eigenvalues are fundamental in solving differential equations involving matrices. They are scalars that give us insight into certain properties of a matrix. To find eigenvalues, for a given matrix \( A \), you solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same size as \( A \).

In the given exercise, solving the characteristic equation leads to the discovery of eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = \lambda_3 = -2 \), the latter being a repeated root. These values play a crucial role in determining the behavior of solutions in systems of differential equations.

• Eigenvalues help classify the stability and the type of motion in dynamical systems.
• Understanding these values aids in predicting how solutions will change over time.
• Repeated eigenvalues may indicate the need for additional consideration, such as the use of Jordan forms.

Eigenvectors

Once we identify the eigenvalues, the next step is finding the eigenvectors, which are non-zero vectors that change only in scale when a linear transformation is applied. For each eigenvalue \( \lambda \), you solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \).

For the exercise in question, this process determined the eigenvectors corresponding to \( \lambda_1 = -1 \) and \( \lambda_2 = \lambda_3 = -2 \). The respective eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 2 \ 0 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \) form a basis for the solution of the system.

Eigenvectors:

• Define the direction of solutions in the vector space.
• When combined with eigenvalues, they construct the general solution to the system.
• Are essential in representing the "mode" of the system's response over time.

Initial Value Problem

An initial value problem (IVP) involves finding a solution to a differential equation that not only satisfies the equation itself but also meets specific initial conditions. In simpler terms, we need to determine the constants of integration using the conditions given at the start.

For our problem, the system of equations derived from the initial conditions \( \mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} \) allows us to solve for the constants \( c_1 \), \( c_2 \), and \( c_3 \). By substituting these initial value conditions into the general solution, we could uniquely determine these constants and, as such, ensure that the solution is both accurate and tailored to the specific scenario being considered.

• IVPs are crucial for unique solutions in real-world scenarios.
• They provide the means to "anchor" the mathematical model to initial conditions.
• Ensures that the solution can predict behavior from a specific starting point.

Differential Equations

Differential equations involve equations containing derivatives which represent rates of change. They are key in modeling a myriad of phenomena in physics, engineering, and other disciplines where change is constant.

The given exercise involved a system of linear differential equations where \( \mathbf{x}' = A \mathbf{x} \). Solving such systems often requires understanding the interplay of rates of change determined through eigenvalues and eigenvectors.

Differential Equations:

• Used to model change, such as decay, growth, and oscillations, dynamically.
• Solutions tend to describe the behavior of an entire system rather than simple numerical values.
• Provide insight into complex phenomena over time.

In this exercise, our task was to find a set of basis solutions (fundamental set of solutions) that satisfactorily describe the behavior of the system with respect to time, anchored by the initial conditions provided.