Question

Repeat Exercise 7 with $$ A=\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right], \quad \mathbf{y}_{1}=\left[\begin{array}{l} e^{3 t} \\ e^{3 t} \end{array}\right], \quad \mathbf{y}_{2}=\left[\begin{array}{r} e^{t} \\ -e^{t} \end{array}\right], \quad \mathbf{k}=\left[\begin{array}{l} 2 \\ 8 \end{array}\right] $$

Short answer

Answer: The particular solution to the nonhomogeneous system is $$\mathbf{y}_{p} = \left[\begin{array}{l} 2t \\ 8t \end{array}\right] + \mathbf{c}$$, and the general solution to the nonhomogeneous system is $$\mathbf{y}(t) = \left[\begin{array}{l} 2t \\ 8t \end{array}\right] + c_{1}\left[\begin{array}{l} e^{3t} \\ e^{3t} \end{array}\right] + c_{2}\left[\begin{array}{r} e^{t} \\ -e^{t} \end{array}\right]$$.

Step-by-step solution

Write down the given information

We are given the matrix A, homogeneous solutions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\), and the vector \(\mathbf{k}\). $$ A=\left[\begin{array}{ll} 2 & 1 \\\ 1 & 2 \end{array}\right], \quad \mathbf{y}_{1}=\left[\begin{array}{l} e^{3 t} \\\ e^{3 t} \end{array}\right], \quad \mathbf{y}_{2}=\left[\begin{array}{r} e^{t} \\\ -e^{t} \end{array}\right], \quad \mathbf{k}=\left[\begin{array}{l} 2 \\\ 8 \end{array}\right] $$

Determine the particular solution

To find the particular solution \(\mathbf{y}_{p}\), we first need to evaluate the vector \((A - 3I)^{-1}\mathbf{k}\). We begin by calculating \(A - 3I\): $$ A - 3I=\left[\begin{array}{ll} 2-3 & 1 \\ 1 & 2-3 \end{array}\right]=\left[\begin{array}{ll} -1 & 1 \\ 1 & -1 \end{array}\right]. $$ Now, we need to find the inverse of \((A - 3I)\): $$ (A - 3I)^{-1} = \frac{1}{\det(A - 3I)} \left[\begin{array}{ll} -1 & -1 \\ -1 & -1 \end{array}\right] = \frac{1}{(-1)^2 - (1)^2} \left[\begin{array}{ll} -1 & -1 \\ -1 & -1 \end{array}\right], $$ where we have used the fact that \(\det(A - 3I)=(-1)^2 - (1)^2=0\). Since the determinant is zero, the inverse does not exist, and hence the method to find the particular solution does not apply here.

Find a different approach for the particular solution

Since the previous method doesn't work, we need to find a different approach to find a particular solution. Since the homogeneous solutions consist of exponential functions, let's try integrating the constant vector \(\mathbf{k}\) in the nonhomogeneous equation \(A\mathbf{y}=\mathbf{k}\): $$ \mathbf{y}_{p} = \int \mathbf{k} dt = \int \left[\begin{array}{l} 2 \\ 8 \end{array}\right] dt = \left[\begin{array}{l} 2t \\ 8t \end{array}\right] + \mathbf{c}, $$ where \(\mathbf{c}\) is an arbitrary constant vector. Now, substitute \(\mathbf{y}_{p}\) into the nonhomogeneous equation \(A\mathbf{y} = \mathbf{k}\): $$ A\mathbf{y}_{p} = \left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} 2t \\ 8t \end{array}\right] + \left[\begin{array}{l} 2 \\ 8 \end{array}\right] = \mathbf{k}. $$ The equation above is satisfied; therefore, $\mathbf{y}_{p} = \left[\begin{array}{l} 2t \\ 8t \end{array}\right] + \mathbf{c}$ is a particular solution.

Find the general solution

Now that we have found a particular solution \(\mathbf{y}_{p}\) to the nonhomogeneous system, we can find the general solution to the nonhomogeneous system by adding the general solution of the homogeneous system to \(\mathbf{y}_{p}\): $$ \mathbf{y}(t) = \mathbf{y}_{p} + c_{1}\mathbf{y}_{1} + c_{2}\mathbf{y}_{2} = \left[\begin{array}{l} 2t \\ 8t \end{array}\right] + c_{1}\left[\begin{array}{l} e^{3t} \\ e^{3t} \end{array}\right] + c_{2}\left[\begin{array}{r} e^{t} \\ -e^{t} \end{array}\right]. $$ This is the general solution to the nonhomogeneous system of equations.

Key concepts

Particular Solution

When dealing with a nonhomogeneous system of differential equations, a particular solution refers to one specific solution that satisfies the nonhomogeneous system. In contrast with homogeneous solutions, which are associated with the associated homogeneous system (the system with zero on the right-hand side), a particular solution enables us to understand how the nonhomogeneity of the system influences the behavior of the solutions.

Finding a particular solution often involves a method of undetermined coefficients or variation of parameters. However, these methods may not always be applicable, such as when the matrix involved does not have an inverse. In the provided exercise, the attempted method to find the particular solution using the matrix inverse was not possible because the determinant of the matrix turned out to be zero, indicating the matrix was not invertible. When this occurs, alternative methods, like using an ansatz tailored to the form of the nonhomogeneous term or integrating if the nonhomogeneous term is a constant, must be used. Here, integrating the constant vector led to a particular solution of the form \(\mathbf{y}_p = \left[\begin{array}{l}2t \ 8t\end{array}\right] + \mathbf{c}\) which satisfied the equation \(A\mathbf{y} = \mathbf{k}\).

Homogeneous Solutions

Moving on to homogeneous solutions, they are solutions to the homogeneous version of the equation, where the nonhomogeneous term (right-hand side) is replaced with a zero vector. It's important to realize that these solutions form a vector space, more specifically, a subspace of the set of all possible solutions to the system of differential equations. They can be combined linearly to produce more solutions, by multiplying them with constants and adding them together.

In our case, the given homogeneous solutions are \(\mathbf{y}_1\) and \(\mathbf{y}_2\), which correspond to exponentials of the form \(e^{3t}\) and \(e^t\). It is critical to ensure that these solutions are indeed linearly independent; otherwise, they would not form a complete basis for the subspace of solutions. When constructing the general solution of a nonhomogeneous equation, these homogeneous solutions are part of the 'homogeneous part' of the answer.

General Solution

The general solution of a nonhomogeneous system is the sum of the particular solution and the general solution of the homogeneous system. To clarify, it embodies all possible solutions to the nonhomogeneous system. The general solution is formulated by adding the particular solution you've found to the linear combination of homogeneous solutions, each multiplied by an arbitrary constant.

In the exercise example, the general solution is expressed as \(\mathbf{y}(t) = \mathbf{y}_p + c_1\mathbf{y}_1 + c_2\mathbf{y}_2\). The constants \(c_1\) and \(c_2\) can be any real numbers and represent the contribution of each homogeneous solution to the overall solution. Homogeneous solutions model the 'unforced' system's behavior, while the particular solution accounts for the influence of the nonhomogeneity. By combining these elements, the general solution can describe the system's state under any initial condition.

Matrix Inverse

Finally, let's delve into the concept of a matrix inverse. The inverse of a matrix \(A\), denoted as \(A^{-1}\), is a matrix that, when multiplied with \(A\), yields the identity matrix \(I\). The identity matrix is the equivalent of '1' in matrix algebra, making \(AA^{-1} = A^{-1}A = I\). In many problems involving systems of linear equations or differential equations, finding an inverse matrix can be a critical step in finding solutions, particularly particular solutions.

However, it's important to note that not all matrices are invertible. A matrix has an inverse only if its determinant is nonzero. In the exercise above, the attempt to find a matrix inverse was thwarted by a zero determinant, which is a definitive sign that the matrix is singular, aka non-invertible. The presence of a non-invertible matrix in the method of finding a particular solution indicates the need to explore alternative strategies to solve the problem at hand.