Question

V\(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}1 & -4 & -1 \\ 3 & 2 & 3 \\ 1 & 1 & 3\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}-1 \\ 9 \\ 4\end{array}\right]\)

Short answer

The eigenvalues of the coefficient matrix are 1, 2, and 3.

Step-by-step solution

Find the eigenvalues of the coefficient matrix

First, we need to find the eigenvalues of the coefficient matrix: \(A = \left[\begin{array}{rrr}1 & -4 & -1 \\\ 3 & 2 & 3 \\\ 1 & 1 & 3\end{array}\right]\) To do this, solve the characteristic equation det\((A - \lambda I) = 0\), where \(I\) is the identity matrix: det\((A - \lambda I) = \left| \begin{array}{ccc} 1-\lambda & -4 & -1 \\ 3 & 2-\lambda & 3 \\ 1 & 1 & 3-\lambda \end{array} \right| = 0\) Computing the determinant, we get: \((-\lambda^3 + 6\lambda^2 - 11\lambda + 6) = 0\) Now we must solve this equation for the eigenvalues \(\lambda\).

Solve the characteristic equation

Factor the equation \((-\lambda^3 + 6\lambda^2 - 11\lambda + 6) = 0\) using factoring techniques, or use a software like WolframAlpha, to find the eigenvalues. We get: \(\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 3\)

Find eigenvectors for each eigenvalue

Next, find the eigenvectors associated with each eigenvalue by solving \((A - \lambda_i I)\mathbf{v}_i = 0\), where \(\mathbf{v}_i\) is the eigenvector corresponding to the eigenvalue \(\lambda_i\): For \(\lambda_1 = 1\): \((A - I)\mathbf{v}_1 = \left[ \begin{array}{rrr} 0 & -4 & -1 \\ 3 & 1 & 3 \\ 1 & 1 & 2\end{array} \right]\mathbf{v}_1 = 0\) \(\implies \mathbf{v}_1 = \left[\begin{array}{r} 1 \\ 0 \\ -1 \\\end{array}\right]\) For \(\lambda_2 = 2\): \((A - 2I)\mathbf{v}_2 = \left[ \begin{array}{rrr} -1 & -4 & -1 \\ 3 & 0 & 3 \\ 1 & 1 & 1\end{array} \right]\mathbf{v}_2 = 0\) \(\implies \mathbf{v}_2 = \left[\begin{array}{r} 1 \\ 1 \\ 2 \\\end{array}\right]\) For \(\lambda_3 = 3\): \((A - 3I)\mathbf{v}_3 = \left[ \begin{array}{rrr} -2 & -4 & -1 \\ 3 & -1 & 3 \\ 1 & 1 & 0\end{array} \right]\mathbf{v}_3 = 0\) \(\implies \mathbf{v}_3 = \left[\begin{array}{r} 1 \\ -1 \\ 1 \\\end{array}\right]\)

Form the general solution

Now form the general solution for \(\mathbf{y}(t)\) using eigenvectors and eigenvalues: \(\mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2 + C_3 e^{\lambda_3 t} \mathbf{v}_3 = C_1 e^{t} \left[\begin{array}{r} 1 \\ 0 \\ -1 \\\end{array}\right] + C_2 e^{2t} \left[\begin{array}{r} 1 \\ 1 \\ 2 \\\end{array}\right] + C_3 e^{3t} \left[\begin{array}{r} 1 \\ -1 \\ 1 \\\end{array}\right]\)

Solve for constants using the initial condition

We are given the initial condition \(\mathbf{y}(0) = \left[\begin{array}{r} -1 \\ 9 \\ 4 \\\end{array}\right]\). Plug in \(t=0\) in the general solution to get: \(\left[\begin{array}{r} -1 \\ 9 \\ 4 \\\end{array}\right] = C_1 \left[\begin{array}{r} 1 \\ 0 \\ -1 \\\end{array}\right] + C_2 \left[\begin{array}{r} 1 \\ 1 \\ 2 \\\end{array}\right] + C_3 \left[\begin{array}{r} 1 \\ -1 \\ 1 \\\end{array}\right]\) Solve for \(C_1\), \(C_2\), and \(C_3\): \(C_1 = 2, \quad C_2 = 1, \quad C_3 = -3\)

Write the final solution

Now plug the constants back into the general solution to get the final solution for \(\mathbf{y}(t)\): \(\mathbf{y}(t) = 2 e^t \left[\begin{array}{r} 1 \\ 0 \\ -1 \\\end{array}\right] + 1 e^{2t} \left[\begin{array}{r} 1 \\ 1 \\ 2 \\\end{array}\right] - 3 e^{3t} \left[\begin{array}{r} 1 \\ -1 \\ 1 \\\end{array}\right]\) This is the solution for the given system of ODEs with the initial condition provided.

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra and play a critical role in the study of differential equations. They are special numbers associated with a square matrix that provide critical information about the behavior of linear transformations and systems of linear equations. When we have a differential equation represented in matrix form, like the given problem \(V\mathbf{y}^{\text{'}} = A\mathbf{y}\), the eigenvalues of the matrix \(A\) tell us about the solutions' growth rate.

In the step-by-step solution, finding the eigenvalues was the first task. To solve for the eigenvalues, we calculated the roots of the characteristic equation, which directly affect the nature of the differential equation's solutions. Different eigenvalues can indicate whether the system's solutions are oscillating, decaying, or growing over time, which is essential in fields such as physics and engineering.

Especially for initial value problems, the eigenvalues play a role in determining the general solution, influencing how the solution evolves from the specified initial condition.

Eigenvectors

Associated with each eigenvalue of a matrix are corresponding eigenvectors, which are non-zero vectors that only change by a scalar factor when that linear transformation is applied. That is, for a matrix \(A\) and an eigenvalue \(\lambda\), an eigenvector \(\mathbf{v}\) satisfies the equation \(A\mathbf{v} = \lambda\mathbf{v}\).

In the context of solving differential equations, once we have determined the eigenvalues, we find the associated eigenvectors as demonstrated in Step 3 of the provided solution. These eigenvectors help us construct the solution to the differential equation. Each eigenvector forms a part of the general solution of the system, granting us a set of basis vectors that describe the direction of the solutions in the phase space.

Eigenvectors are essential in understanding the behavior of solutions to linear systems since they dictate the structure of the solution space. They describe the principal directions in which the associated linear transformation stretches or compresses the space.

Characteristic Equation

The characteristic equation is a polynomial equation derived from a matrix which is vital in determining the eigenvalues. For a given square matrix \(A\), the characteristic equation is formulated as det\((A - \lambda I) = 0\), where \(\lambda\) represents an eigenvalue and \(I\) is the identity matrix of the same dimension as \(A\). Solving this equation reveals the eigenvalues.

In the solution process, after establishing the matrix \(A\) from the differential equation, we calculated the determinant of \(A - \lambda I\) to get the characteristic equation. By finding the roots of this polynomial, we obtained the system's eigenvalues. These are crucial steps as the eigenvalues and characteristic equation together provide deep insight into the stability and dynamics of a system described by a set of differential equations.

Initial Value Problems

Initial Value Problems (IVPs) are a type of differential equation where the solution path is specified at a particular point, known as the initial value. In the context of systems of linear differential equations, the initial value sets a condition that the solution must satisfy at a certain time, typically \(t = 0\).

The final steps of our solution, Steps 5 and 6, involve using the initial value \(\mathbf{y}(0)\) to find the specific constants for our general solution. By substituting \(t = 0\) into the general solution and equating it to the given initial value, we get a system of equations that allows us to solve for the constants \(C_1, C_2, \)and\( C_3\). These constants are then plugged back into the general solution to get the particular solution that not only solves the differential equation but also satisfies the given initial condition.

Understanding IVPs is crucial for applying differential equations to real-world scenarios, where we often have information about the state of a system at a specific time and are interested in predicting its future behavior.