Question

In each exercise, the coefficient matrix \(A\) of the given linear system has a full set of eigenvectors and is therefore diagonalizable. (a) As in Example 4 , make the change of variables \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), where \(T^{-1} A T=D\). Reformulate the given problem as a set of uncoupled problems. (b) Solve the uncoupled system in part (a) for \(\mathbf{z}(t)\), and then form \(\mathbf{y}(t)=T \mathbf{z}(t)\) to obtain the solution of the original problem.\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}-4 & -6 \\ 3 & 5\end{array}\right] \mathbf{y}+\left[\begin{array}{r}e^{2 t} \\ -e^{2 t}\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}0 \\\ 0\end{array}\right]\)

Short answer

Question: Find the solution to the linear system of differential equations with the initial condition y(0) = 0. Answer: The solution to the original problem is y(t) = (2/5)e^(2t) - 1/5)e^(-t), -1/5)e^(2t) + 1/5)e^(-t).

Step-by-step solution

Find eigenvalues and eigenvectors of A

The given matrix A is \(A = \begin{bmatrix} -4 & -6 \\ 3 & 5 \end{bmatrix}\) To find the eigenvalues, compute the characteristic equation: \(\det(A - \lambda I) = 0\) By calculating: \((\begin{bmatrix} -4-\lambda & -6 \\ 3 & 5-\lambda \end{bmatrix}) = (5-\lambda)(-4-\lambda)-(-6)(3) = \lambda^2 + \lambda - 2\) Solving the quadratic equation, we get λ = -1 and λ = 2. Now compute eigenvectors for each eigenvalue: For λ= -1: \((A + I) \begin{bmatrix} x \\ y \end{bmatrix} = 0\), where \((A + I) = \begin{bmatrix} -3 & -6 \\ 3 & 6 \end{bmatrix}\) Row reducing, the matrix becomes \(\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}\), so the eigenvector is \(\begin{bmatrix} -2 \\ 1\end{bmatrix}\). For λ= 2: \((A - 2I) \begin{bmatrix} x \\ y \end{bmatrix} = 0\), where \((A-2I) = \begin{bmatrix} -6 & -6 \\ 3 & 3 \end{bmatrix}\) Row reducing, the matrix becomes \(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}\), so the eigenvector is \(\begin{bmatrix} 1 \\ -1\end{bmatrix}\).

Diagonalize A

We create the diagonal matrix D with eigenvalues and T with eigenvectors: \(D = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\) and \(T = \begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix}\) Next, calculate \(T^{-1}\): \(T^{-1} = \frac{1}{(-2)(-1)-1(1)}\begin{bmatrix} -1 & -1 \\ -1 & -2 \end{bmatrix} =\begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}\)

Make change of variables and reformulate uncoupled problems

Make the change of variables as \(\mathbf{z}(t) = T^{-1} \mathbf{y}(t)\). This yields \(T\mathbf{z}'(t) = \left[\begin{array}{rr}-4 & -6 \\\ 3 & 5\end{array}\right] T\mathbf{z}(t)+\left[\begin{array}{r}e^{2 t} \\\ -e^{2 t}\end{array}\right]\). Now multiply both sides by \(T^{-1}\) to obtain: \(\mathbf{z}'(t) = T^{-1} A T \mathbf{z}(t)+T^{-1}\left[\begin{array}{r}e^{2 t} \\\ -e^{2 t}\end{array}\right]\) This becomes: \(\mathbf{z}'(t) = D\mathbf{z}(t)+\left[\begin{array}{r}\frac{2}{3} e^{2 t} \\ 0\end{array}\right]\)

Solve the uncoupled system for z(t)

The system is now uncoupled and can be solved as two separate first-order linear differential equations: \(z_1'(t) = -z_1(t) + \frac{2}{3}e^{2t}\) with \(z_1(0) = 0\) \(z_2'(t) = 2z_2(t)\) with \(z_2(0) = 0\) Solving these equations, we get: \(z_1(t) = -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t}\) \(z_2(t) = 0\)

Find the solution for the original problem y(t)

Since we have \(\mathbf{z}(t) = \begin{bmatrix} -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t}\\ 0 \end{bmatrix}\), we can compute \(\mathbf{y}(t) = T\mathbf{z}(t)\): \(\mathbf{y}(t) = \begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{2}{5}e^{2t} - \frac{1}{5}e^{-t} \\ -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t} \end{bmatrix}\) Hence, the solution to the original problem is \(\mathbf{y}(t) = \begin{bmatrix} \frac{2}{5}e^{2t} - \frac{1}{5}e^{-t} \\ -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t} \end{bmatrix}\).

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is critical when diagonalizing matrices or solving differential equations. An eigenvalue \(\lambda\) for a matrix \(A\) is a number such that there exists a non-zero vector, known as an eigenvector, satisfying the equation: \(A\mathbf{v} = \lambda \mathbf{v}\). This means that when \(A\) acts on an eigenvector \(\mathbf{v}\), it simply rescales the vector by the factor \(\lambda\), the eigenvalue.
To find the eigenvalues, solve the characteristic equation obtained by setting the determinant of \(A - \lambda I\) to zero, where \(I\) is the identity matrix. In our example where \(A = \begin{bmatrix} -4 & -6 \ 3 & 5 \end{bmatrix}\), the characteristic equation is \(\lambda^2 + \lambda - 2 = 0\), yielding eigenvalues \(\lambda = -1\) and \(\lambda = 2\).
Once eigenvalues are determined, eigenvectors are found by solving \((A - \lambda I) \mathbf{v} = \mathbf{0}\) for each eigenvalue. Each eigenvalue/eigenvector pair contributes significantly to understanding the geometric properties of the matrix \(A\) and ultimately aids in transforming a system into a simpler form.

Uncoupled System of Differential Equations

Once we've identified the eigenvalues and eigenvectors of a matrix, we can diagonalize the matrix to simplify our system of differential equations. By using the change of variables \(\mathbf{z}(t) = T^{-1} \mathbf{y}(t)\), where \(T\) is the matrix formed by eigenvectors, we convert the system into one described by a diagonal matrix \(D\).
In our case, \(T = \begin{bmatrix} -2 & 1 \ 1 & -1 \end{bmatrix}\) and \(T^{-1} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix}\). This transformation leads to an uncoupled system \(\mathbf{z}'(t) = D \mathbf{z}(t) + \text{other terms}\) where each equation in the system only involves itself. For instance, \[ \mathbf{z}'(t) = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix}\mathbf{z}(t) + \begin{bmatrix} \frac{2}{3} e^{2t} \ 0 \end{bmatrix}\] results in two separate equations. Solving an uncoupled system is generally easier as each equation can be tackled independently.

Solution of Initial Value Problems

After achieving an uncoupled form, solving the differential equations becomes a tractable task, aided significantly by initial conditions. An initial value problem involves solving a differential equation by using specific starting values. In our example \(\mathbf{z}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), the initial conditions give specific solutions to the uncoupled equations.
For instance, \(z_1'(t) = -z_1(t) + \frac{2}{3}e^{2t}\) with \(z_1(0) = 0\) is solved to yield \(z_1(t) = -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t}\). Similarly, \(z_2'(t) = 2z_2(t)\) with \(z_2(0) = 0\) simplifies further to \(z_2(t) = 0\).
Once \(\mathbf{z}(t)\) is determined, the original system's solution \(\mathbf{y}(t)\) is retrieved by transforming back: \(\mathbf{y}(t) = T \mathbf{z}(t)\). For this case, it leads us to the final solution \(\mathbf{y}(t) = \begin{bmatrix} \frac{2}{5}e^{2t} - \frac{1}{5}e^{-t} \ -\frac{1}{5}e^{2t} + \frac{1}{5}e^{-t} \end{bmatrix}\), showcasing how initial conditions guide the complete resolution of a differential equation system.