Question

In Exercises \(1-10\) find a particular solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -4 & -3 \\ 6 & 5 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} 2 \\ -2 e^{t} \end{array}\right] $$

Short answer

Answer: \(\mathbf{y_p}(t) = \begin{bmatrix}2 \\ 1 \end{bmatrix} e^t\)

Step-by-step solution

Assume a particular solution of the form \(\mathbf{y_p}(t) = C e^{rt}\)

Let's assume that the particular solution for the given linear system of differential equations is in the form: $$ \mathbf{y_p}(t) = C e^{rt} $$ with \(C = \begin{bmatrix}c_1 \\ c_2 \end{bmatrix}\) being a constant vector.

Calculate the derivative of \(\mathbf{y_p}(t)\)

Now, let's calculate the derivative of the assumed particular solution with respect to time \(t\): $$ \frac{d\mathbf{y_p}(t)}{dt} = r C e^{rt} $$

Substitute the assumed particular solution and its derivative in the given equation

Next, substitute \(\mathbf{y_p}(t)\) and its derivative in the given equation: $$ rCe^{rt} = A Ce^{rt} + \begin{bmatrix}2 \\ -2e^t \end{bmatrix} $$ Since \(e^{rt}\) is nonzero, we can divide by it: $$ rC = AC + \begin{bmatrix}2 \\ -2e^{(1-r)t} \end{bmatrix} $$

Solve for the constant vector \(C\)

Now, let's solve for the constant vector \(C\). Notice that the second component of the matrix on the right side depends on \(e^{(1-r)t}\), which means we must choose \(r=1\). This leads to the following system of equations: $$ \begin{bmatrix}r-(-4) & -3 \\ 6 & r-5 \end{bmatrix}C = \begin{bmatrix}2 \\ -2 \end{bmatrix} $$ Substitute \(r=1\) and solve for \(C\): $$ \begin{bmatrix}5 & -3 \\ 6 & -4 \end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix}2 \\ -2 \end{bmatrix} $$ To solve this system of equations, we can use the inverse matrix method: $$ \begin{bmatrix}c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix}5 & -3 \\ 6 & -4 \end{bmatrix}^{-1} \begin{bmatrix}2 \\ -2 \end{bmatrix} $$ By finding the inverse of the matrix and multiplying it with the given vector: $$ C = \begin{bmatrix}c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix}2 \\ 1 \end{bmatrix} $$

Write down the particular solution

Finally, substitute the obtained constant vector \(C\) and the chosen eigenvalue \(r=1\) into the assumed particular solution expression: $$ \mathbf{y_p}(t) = \begin{bmatrix}2 \\ 1 \end{bmatrix} e^t $$ This is the particular solution for the given linear system of differential equations.

Key concepts

Linear System of Differential Equations

When faced with a system of differential equations, one is essentially dealing with a set of equations that describe how variables change with respect to one another. In the context of a linear system of differential equations, these changes are described using linear combinations of the unknown functions and their derivatives.

A standard form of a linear system of differential equations is represented as \[\begin{equation}\mathbf{y}' = A\mathbf{y} + \mathbf{B}(t),\end{equation}\]where \(\mathbf{y}\) is a vector of unknown functions, \(A\) is a matrix of coefficients, and \(\mathbf{B}(t)\) is a vector of functions for nonhomogeneous terms. The goal is to find a function \(\mathbf{y}(t)\) that satisfies the system for all values of \(t\).

To determine a particular solution, one must often make an educated guess about the form of \(\mathbf{y}_p(t)\) based on the nonhomogeneous component \(\mathbf{B}(t)\). This approach, known as the method of undetermined coefficients, involves plugging the assumed solution into the system and solving for the coefficients that make the equation work for all \(t\).

Eigenvalues and Eigenvectors

The concepts of eigenvalues and eigenvectors play a vital role in solving linear systems of differential equations. An eigenvalue is a scalar that indicates how an associated eigenvector is scaled when it is multiplied by a given matrix. Let's consider a matrix \(A\). If there's a vector \(\mathbf{v}\) and a scalar \(\lambda\) such that\[\begin{equation}A\mathbf{v} = \lambda\mathbf{v},\end{equation}\]then \(\mathbf{v}\) is an eigenvector of \(A\), and \(\lambda\) is the corresponding eigenvalue.

For systems of differential equations, eigenvalues can indicate the type of growth, decay, or oscillations solutions may exhibit, while eigenvectors can give the direction in which these behaviors occur. Calculating them involves finding the roots of the characteristic polynomial obtained from the equation\[\begin{equation}\text{det}(A - \lambda I) = 0,\end{equation}\]where \(I\) is the identity matrix of the same size as \(A\). The eigenvalues are then used to facilitate the process of finding particular solutions to the system of differential equations. Importantly, when matching terms for nonhomogeneous functions, the eigenvalues help in determining the appropriate form of the solution.

Matrix Method for Solving Systems

The matrix method for solving systems of linear equations is a powerful tool encapsulated largely within linear algebra. It relies heavily on the use of matrices and their inverses to find solutions. For linear systems represented by \[\begin{equation}A\mathbf{x} = \mathbf{b},\end{equation}\]where \(A\) is a square matrix of coefficients, \(\mathbf{x}\) is the vector of unknowns, and \(\mathbf{b}\) is the vector of constants, solving for \(\mathbf{x}\) can be done by finding the inverse of matrix \(A\) and performing matrix multiplication:

\[\begin{equation}\mathbf{x} = A^{-1}\mathbf{b}.\end{equation}\]This method can only be applied when \(A\) is non-singular, meaning it has an inverse. To calculate the inverse of a matrix, one might use techniques such as the adjugate method, Gaussian elimination, or leverage software tools that are built for such calculations. Once the inverse is found, the solution is straightforward to compute. In the context of differential equations solutions, the matrix method simplifies the process of finding the particular integral when the system is linear.

In the given problem, students are asked to find a particular solution by assuming a form for \(\mathbf{y}_p(t)\), calculating its derivative, substituting into the system, and using the matrix method to solve the resulting system of equations. Identifying the correct form for the particular solution and accurately applying the matrix method are critical for solving such problems.