Question

Solve the initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}3 & 0 & 1 \\ 11 & -2 & 7 \\ 1 & 0 & 3\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}2 \\ 7 \\ 6\end{array}\right]\)

Short answer

In summary, the unique solution to the given initial value problem is: $$ \mathbf{y}(t)=7\begin{bmatrix}0\\1\\0\end{bmatrix} + 6\exp(2t)\begin{bmatrix}1\\1/2\\0\end{bmatrix} + 4\exp(-t)\begin{bmatrix}-1\\-10\\1\end{bmatrix} $$

Step-by-step solution

Diagonalize the matrix

First, let's find the eigenvalues of the matrix \(A = \left[\begin{array}{rrr}3 & 0 & 1 \\\ 11 & -2 & 7 \\\ 1 & 0 & 3\end{array}\right]\). The characteristic polynomial is given by: $$ \det{(A-\lambda I)} = \begin{vmatrix} 3-\lambda & 0 & 1 \\\ 11 & -2-\lambda & 7 \\\ 1 & 0 & 3-\lambda \end{vmatrix} $$ Compute the determinant to get the characteristic polynomial: $$ (3-\lambda)((-2-\lambda)(3-\lambda)-0)-11(7(3-\lambda)) + (1)(7)\\ \Rightarrow-\lambda^3+4\lambda^2-(-22\lambda)+6\lambda^2 \Rightarrow-\lambda^3+4\lambda^2+\lambda^2 $$ This polynomial factors as \(-\lambda^3+4\lambda^2+\lambda^2=-\lambda(\lambda-2)(\lambda+1)\). Therefore, the eigenvalues are \(\lambda_1=0\), \(\lambda_2=2\), and \(\lambda_3=-1\). Next, for each eigenvalue, find an eigenvector: - For \(\lambda_1=0\), solve the equation \((A-0I)v_1=0\) for \(v_1\), - For \(\lambda_2=2\), solve the equation \((A-2I)v_2=0\) for \(v_2\), - For \(\lambda_3=-1\), solve the equation \((A+1I)v_3=0\) for \(v_3\). After solving these equations, we get the eigenvectors \(v_1=\begin{bmatrix}0\\1\\0\end{bmatrix}\), \(v_2=\begin{bmatrix}1\\1/2\\0\end{bmatrix}\), and \(v_3=\begin{bmatrix}-1\\-10\\1\end{bmatrix}\). Now, let's form the matrix \(P\) with eigenvectors as columns and compute its inverse \(P^{-1}\): $$ P=\left[\begin{array}{rrr}0 & 1& -1\\1 & 1/2 & -10\\0 & 0 & 1\end{array}\right], $$ and $$ P^{-1}=\left[\begin{array}{rrr}0 & 2 & 1\\1 & -2 & -1\\0 & 0 & 1\end{array}\right]. $$

Solve the system of linear equations

Now, using \(P\) and \(P^{-1}\), we can diagonalize the matrix \(A\): $$ D=P^{-1}AP = \left[\begin{array}{rrr}0 & 0 & 0\\0 & 2 & 0\\0 & 0 & -1\end{array}\right] $$ We now can find the general solution to the system of linear equations: $$ \mathbf{y}(t) = P\begin{bmatrix}c_1\exp(\lambda_1 t)\\c_2\exp(\lambda_2 t)\\c_3\exp(\lambda_3 t)\end{bmatrix} = P\begin{bmatrix}c_1\\c_2\exp(2t)\\c_3\exp(-t)\end{bmatrix} $$ Substituting the expressions for \(P\), \(v_1\), \(v_2\), and \(v_3\), we get: $$ \mathbf{y}(t)=\left[\begin{array}{rrr}0 & 1& -1\\1 & 1/2 & -10\\0 & 0 & 1\end{array}\right]\begin{bmatrix}c_1\\c_2\exp(2t)\\c_3\exp(-t)\end{bmatrix} \\ = c_1\begin{bmatrix}0\\1\\0\end{bmatrix} + c_2\exp(2t)\begin{bmatrix}1\\1/2\\0\end{bmatrix} + c_3\exp(-t)\begin{bmatrix}-1\\-10\\1\end{bmatrix} $$

Apply the initial condition

Finally, let's plug in the initial value \(\mathbf{y}(0)=\left[\begin{array}{l}2 \\\ 7 \\\ 6\end{array}\right]\) to find the constants \(c_1\), \(c_2\), and \(c_3\): $$ \mathbf{y}(0)=c_1\begin{bmatrix}0\\1\\0\end{bmatrix} + c_2\begin{bmatrix}1\\1/2\\0\end{bmatrix} + c_3\begin{bmatrix}-1\\-10\\1\end{bmatrix} = \left[\begin{array}{l}2 \\\ 7 \\\ 6\end{array}\right] $$ Solving this system of linear equations, we get \(c_1=7\), \(c_2=6\), and \(c_3=4\). Therefore, the unique solution to the initial value problem is: $$ \mathbf{y}(t)=7\begin{bmatrix}0\\1\\0\end{bmatrix} + 6\exp(2t)\begin{bmatrix}1\\1/2\\0\end{bmatrix} + 4\exp(-t)\begin{bmatrix}-1\\-10\\1\end{bmatrix} $$

Key concepts

Matrix Diagonalization

Matrix diagonalization is a process used to simplify complex matrices for easier mathematical operations. In the context of differential equations, particularly systems of linear equations, diagonalization simplifies solving these systems by transforming the original matrix into a diagonal matrix. A diagonal matrix is a matrix where entries outside the main diagonal are zero. Diagonal matrices are beneficial because their power is easy to calculate, which helps in solving systems of differential equations.

To diagonalize a matrix, you need to find a matrix of eigenvectors, denoted as \( P \), and its inverse, \( P^{-1} \). The transformation is represented as \( D = P^{-1}AP \), where \( A \) is the original matrix, and \( D \) is the diagonal matrix consisting of eigenvalues on its diagonal. This step reduces the complexity of matrices in differential equations.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in linear algebra, particularly for solving differential equations. An eigenvalue is a scalar that indicates how much the direction of an eigenvector is stretched during transformation by a matrix. On the other hand, an eigenvector is a non-zero vector that only changes by the scalar factor during this transformation.

To find eigenvalues of a matrix, you compute the determinant of \( A - \lambda I = 0 \), where \( A \) is the matrix and \( I \) is the identity matrix. The solutions to this characteristic polynomial are the eigenvalues. After finding eigenvalues, we solve \( (A - \lambda I)v = 0 \) to find the respective eigenvectors \( v \) for each eigenvalue \( \lambda \). These help in constructing the matrix \( P \) used in diagonalization, where \( P \) is composed of column vectors corresponding to each eigenvector.

Initial Value Problems

Initial value problems in differential equations involve finding a specific solution given an initial condition or starting point. It's crucial in many real-world applications, such as physics, engineering, and economics, where the state of a system at a particular time needs to be defined.

In an initial value problem, you solve a differential equation and apply the initial conditions to determine the exact values of any constants involved in your general solution. For example, if you have a system \( \mathbf{y}' = A \mathbf{y} \) with \( \mathbf{y}(0) = \begin{bmatrix} a \ b \ c \end{bmatrix} \), you solve the system generally and then adjust your solution to satisfy \( \mathbf{y}(0) \).

The initial values provide specific information that can be plugged into the general solution to solve for constants or coefficients, resulting in the precise solution needed for the scenario.

Systems of Linear Equations

Systems of linear equations are sets of equations with multiple variables. Solving them involves finding the variable values that satisfy all equations simultaneously. In differential equations, these systems often appear in modeling multiple interacting quantities that depend on one another.

To solve systems, especially in a matrix form like \( A\mathbf{y} = \mathbf{b} \), you can use matrix operations or methods such as substitution or elimination. When addressing a system arising from differential equations, solving often requires finding eigenvectors and eigenvalues first. This makes diagonalization feasible, turning our problem into something more manageable.

Once you have diagonalized the matrix, finding the solution involves using linear combinations of exponential functions of time, adjusted by eigenvectors. Solving such systems provides the framework to describe how complex dynamic systems evolve over time under given conditions.