Question

(a) Rewrite the given system of linear homogeneous differential equations as a homogeneous linear system of the form \(\mathbf{y}^{\prime}=P(t) \mathbf{y}\). (b) Verify that the given function \(\mathbf{y}(t)\) is a solution of \(\mathbf{y}^{\prime}=P(t) \mathbf{y}\). $$ \begin{aligned} &y_{1}^{\prime}=-3 y_{1}-2 y_{2} \\ &y_{2}^{\prime}=4 y_{1}+3 y_{2} \end{aligned}, \quad \mathbf{y}(t)=\left[\begin{array}{c} e^{t}+e^{-t} \\ -2 e^{t}-e^{-t} \end{array}\right] $$

Short answer

Answer: The equivalent homogeneous linear system is \(\mathbf{y}^{\prime}=\left[\begin{array}{cc} -3 & -2 \\ 4 & 3 \end{array}\right] \mathbf{y}\), and yes, the given function \(\mathbf{y}(t)\) is a solution to this system.

Step-by-step solution

Rewrite the system as a homogeneous linear system

The given system of linear homogeneous differential equations is: $$ \begin{aligned} &y_{1}^{\prime}=-3 y_{1}-2 y_{2} \\\ &y_{2}^{\prime}=4 y_{1}+3 y_{2} \end{aligned} $$ Recall that a homogeneous linear system of the form \(\mathbf{y}^{\prime}=P(t) \mathbf{y}\) is given by: $$\mathbf{y}^{\prime}=\left[\begin{array}{cc} a & b \\\ c & d \end{array}\right] \mathbf{y}$$ Comparing the given system and the form above, we can easily find that: $$P(t)=\left[\begin{array}{cc} -3 & -2 \\\ 4 & 3 \end{array}\right]$$ So, the homogeneous linear system is: $$\mathbf{y}^{\prime}=\left[\begin{array}{cc} -3 & -2 \\\ 4 & 3 \end{array}\right] \mathbf{y}$$

Verify that the given function is a solution

We are given the function \(\mathbf{y}(t)\) as: $$\mathbf{y}(t)=\left[\begin{array}{c} e^{t}+e^{-t} \\\ -2 e^{t}-e^{-t} \end{array}\right]$$ To verify that this function is a solution of the homogeneous linear system, we need to compute its derivative and check if it satisfies the equation \(\mathbf{y}^{\prime}=P(t) \mathbf{y}\). Calculate the derivative of \(\mathbf{y}(t)\): $$\mathbf{y}^{\prime}(t)=\left[\begin{array}{c} e^{t}-e^{-t} \\\ -2 e^{t}+e^{-t} \end{array}\right]$$ Now, let's calculate \(P(t) \mathbf{y}\): $$P(t) \mathbf{y}=\left[\begin{array}{cc} -3 & -2 \\\ 4 & 3 \end{array}\right] \left[\begin{array}{c} e^{t}+e^{-t} \\\ -2 e^{t}-e^{-t} \end{array}\right]=\left[\begin{array}{c} e^{t}-e^{-t} \\\ -2 e^{t}+e^{-t} \end{array}\right]$$ Since \(\mathbf{y}^{\prime}(t)=P(t) \mathbf{y}\), the given function \(\mathbf{y}(t)\) is indeed a solution of the homogeneous linear system.

Key concepts

Linear Systems of Differential Equations

In the realm of calculus, a linear system of differential equations consists of multiple equations linking several functions and their derivatives. Such systems arise frequently in various scientific fields, including physics, engineering, and economics, where multiple interconnected quantities change over time.

The fundamental structure of a linear system of differential equations in two variables, which can be extended to more, is given by \[\begin{aligned}&y_1'(t) = a_{11}y_1(t) + a_{12}y_2(t) + \ &y_2'(t) = a_{21}y_1(t) + a_{22}y_2(t)\end{aligned}\]Here, the primes denote derivatives with respect to time, and the coefficients \(a_{ij}\) are generally functions of time, defining how each function \(y_i(t)\) affects the rate of change of the others. To solve such a system, students must often transform it into matrix form, which simplifies the analysis and solution of the system.

Homogeneous Solutions

The term 'homogeneous' in the context of linear differential equations describes a system where the output is directly proportional to the input, and there are no external influences or non-zero constants. Essentially, homogeneous systems echo the principle of superposition, where the sum of solutions is also a solution.

In mathematical terms, a linear differential equation, or system of equations, is homogeneous if it can be written as \[\mathbf{y}'(t) = A\mathbf{y}(t)\]where \(\mathbf{y}(t)\) is the vector of functions we are trying to find, \(A\) is a matrix of coefficients, and \(\mathbf{y}'(t)\) is the vector of derivatives. A solution to this system is a set of functions that satisfy the equation for all values of \(t\). Homogeneous systems have a key property: if \(\mathbf{y}(t)\) is a solution, then any scalar multiple \(c\cdot\mathbf{y}(t)\) is also a solution. This characteristic is fundamental when combining solutions to compose a general solution for the system.

Matrix Method for Differential Equations

The matrix method provides a powerful tool for solving linear systems of differential equations. This method transforms the system into a matrix equation that can be manipulated and solved using linear algebra techniques.

To apply the matrix method, one must arrange the system into a compact form: \[\mathbf{y}'(t) = P(t)\mathbf{y}(t)\]where \(P(t)\) is a matrix of coefficients. For a simple two-variable system, this matrix is a \(2 \times 2\) matrix, but larger systems will have correspondingly larger matrices. One of the advantages of this approach is that it is often simpler to analyze and solve matrix equations compared to the corresponding system of differential equations. For stationary systems, where the matrix \(P(t)\) is constant, techniques like eigenvalue decomposition can help find explicit solutions.

In practice, the matrix method allows us to treat a complicated system of differential equations as a singular entity, bringing the vast toolbox of linear algebra to our aid. The elegance and efficiency of this method make it a cornerstone for students and professionals when addressing linear differential equations systems.