Question

Consider the linear system $$ \begin{aligned} &x^{\prime}=5 x+3 y \\ &y^{\prime}=4 x+y \end{aligned} $$ (a) Show that $$ x=3 e^{7 t}, \quad x=e^{-t}, $$ and $$ y=2 e^{7 t}, \quad y=-2 e^{-t} $$ are solutions of this system. (b) Show that the two solutions of part (a) are linearly independent on every interval \(a \leq t \leq b\), and write the general solution of the system. (c) Find the solution $$ \begin{aligned} &x=f(t) \\ &y=g(t), \end{aligned} $$ of the system which is such that \(f(0)=0\) and \(g(0)=8\).

Short answer

The given functions are indeed solutions of the linear system, and they are linearly independent as their Wronskian is non-zero. The general solution to the linear system is given by a linear combination of the individual solutions: \[ \begin{aligned} x(t) &= C_1(3e^{7t}) + C_2(e^{-t}) \\ y(t) &= C_1(2e^{7t}) + C_2(-2e^{-t}) \end{aligned} \] Using the given initial conditions \(f(0) = 0\) and \(g(0) = 8\), we find that the constants are \(C_1 = 2\) and \(C_2 = -6\). Thus, the particular solution of the given linear system is: \[ \begin{aligned} f(t) &= 6e^{7t} - 6e^{-t} \\ g(t) &= 4e^{7t} + 12e^{-t} \end{aligned} \]

Step-by-step solution

Verify the Given Functions as Solutions of the System

The given system of linear equations is: \[ \begin{aligned} x' &= 5x + 3y \\ y' &= 4x + y \end{aligned} \] The functions given are \(x_1(t) = 3e^{7t}\), \(x_2(t) = e^{-t}\), \(y_1(t) = 2e^{7t}\), and \(y_2(t) = -2e^{-t}\). Let's calculate the derivatives of these functions with respect to time 't'. \[ \begin{aligned} x_1'(t) &= 21e^{7t} \\ x_2'(t) &= -e^{-t} \\ y_1'(t) &= 14e^{7t} \\ y_2'(t) &= 2e^{-t} \end{aligned} \] Now, let's substitute these functions and their derivatives into the original system of equations and check if they hold true. For \(x_1(t)\) and \(y_1(t)\): \[ \begin{aligned} x_1'(t) &= 5x_1(t) + 3y_1(t) \\ 21e^{7t} &= 5(3e^{7t}) + 3(2e^{7t}) \\ 21e^{7t} &= 21e^{7t} \end{aligned} \] \[ \begin{aligned} y_1'(t) &= 4x_1(t) + y_1(t) \\ 14e^{7t} &= 4(3e^{7t}) + 2e^{7t} \\ 14e^{7t} &= 14e^{7t} \end{aligned} \] For \(x_2(t)\) and \(y_2(t)\): \[ \begin{aligned} x_2'(t) &= 5x_2(t) + 3y_2(t) \\ -e^{-t} &= 5(e^{-t}) + 3(-2e^{-t}) \\ -e^{-t} &= -e^{-t} \end{aligned} \] \[ \begin{aligned} y_2'(t) &= 4x_2(t) + y_2(t) \\ 2e^{-t} &= 4(e^{-t}) - 2e^{-t} \\ 2e^{-t} &= 2e^{-t} \end{aligned} \] The given functions are indeed solutions of the linear system.

Show Linear Independence and Find the General Solution

For the given solutions to be linearly independent, the Wronskian of the functions should be non-zero. The Wronskian for this pair of solutions is given by: \[ W(x_1, x_2) = \begin{vmatrix} x_1 & x_2 \\ y_1 & y_2 \\ \end{vmatrix} = (3e^{7t})(-2e^{-t}) - (e^{-t})(2e^{7t}) \] \[ W(x_1, x_2) = -6e^{6t} - 2e^{6t} = -8e^{6t} \] Since the Wronskian is non-zero, the two solutions are linearly independent. The general solution to the linear system is given by a linear combination of the individual solutions: \[ \begin{aligned} x(t) &= C_1x_1(t) + C_2x_2(t) = C_1(3e^{7t}) + C_2(e^{-t}) \\ y(t) &= C_1y_1(t) + C_2y_2(t) = C_1(2e^{7t}) + C_2(-2e^{-t}) \end{aligned} \] Where \(C_1\) and \(C_2\) are constants to be determined by the initial conditions.

Find the Particular Solution with Given Initial Conditions

The initial conditions are given as: \[ \begin{aligned} f(0) &= 0 \\ g(0) &= 8 \end{aligned} \] Let's plug these values into our general solution: \[ \begin{aligned} 0 &= C_1(3e^{0}) + C_2(e^{0}) \\ 8 &= C_1(2e^{0}) + C_2(-2e^{0}) \end{aligned} \] By simplifying the equations, we get: \[ \begin{aligned} 0 &= 3C_1 + C_2 \\ 8 &= 2C_1 - 2C_2 \end{aligned} \] Solving this system of equations, we find that \(C_1 = 2\) and \(C_2 = -6\). Therefore, the particular solution of the given linear system is: \[ \begin{aligned} x(t) &= 2(3e^{7t}) - 6(e^{-t}) = 6e^{7t} - 6e^{-t} \\ y(t) &= 2(2e^{7t}) - 6(-2e^{-t}) = 4e^{7t} + 12e^{-t} \end{aligned} \] The final solution is: \[ \begin{aligned} f(t) &= 6e^{7t} - 6e^{-t} \\ g(t) &= 4e^{7t} + 12e^{-t} \end{aligned} \]

Key concepts

Homogeneous Systems

In the world of differential equations, a homogeneous system refers to a set of linear differential equations where the right-hand side is zero. This kind of system often takes the form:

• \( x' = Ax \)
• \( y' = By \)

This essentially means that every term in the equation is related to either the variable itself or its derivatives, and there are no constant or standalone terms independent of the functions.
These systems are pivotal in analyzing the behavior of linear systems as they provide insight into the natural state of the system without external or additional influences. Homogeneous systems are crucial for understanding simple dynamic models where external forces do not play a role. For the given exercise with functions like \( x = 3e^{7t} \) and \( y = 2e^{7t} \), we see that both derivatives of \(x\) and \(y\) depend solely on the terms formed by multiplying constant coefficients with \(x\) and \(y\) themselves.

Linear Independence

Linear independence is a critical concept in linear algebra and differential equations. It describes a set of functions where no function can be written as a linear combination of others. In simpler terms, each function provides unique information that cannot be derived from the others.
To determine the linear independence of solutions such as those described in the original exercise, we rely on calculating the Wronskian. If the Wronskian is non-zero, then the functions are linearly independent.
Knowing that solutions are independent helps in forming the general solution to the differential equation. For example, if \(x_1 = 3e^{7t}\) and \(x_2 = e^{-t}\) are linearly independent, it confirms that a valid general solution can be formed as a linear combination of these functions with arbitrary constants, providing a complete description of the system's potential behavior.

Wronskian

The Wronskian is a determinant used to assess the linear independence of a set of functions. Given functions \(y_1, y_2, ..., y_n\), the Wronskian \(W(y_1, y_2, ..., y_n)\) is constructed as follows:

• Each function forms a row in a matrix, with its derivatives filling out subsequent rows.
• The determinant of this matrix is the Wronskian.

In the exercise, to evaluate the independence of functions \(3e^{7t}\), \(e^{-t}\), \(2e^{7t}\), and \(-2e^{-t}\), their Wronskian is computed. Here, it is derived and found to be non-zero, specifically \(-8e^{6t}\), indicating that the functions are independent. The calculation involves determinant properties and directly confirms the necessity of the Wronskian in solving differential equation systems.

Initial Conditions

Initial conditions provide specific values for function solutions at a particular point, crucial for determining unique solutions in differential equations. They essentially "anchor" the solution to a specific starting point and refine the infinite possibilities presented by a general solution.
In the given system, we had initial conditions specified as \(f(0) = 0\) and \(g(0) = 8\). By applying these conditions to the general solution, you can solve for the constants within the linear combination, yielding particular solutions such as \(x(t) = 6e^{7t} - 6e^{-t}\) and \(y(t) = 4e^{7t} + 12e^{-t}\). Initial conditions ensure that the solution is not just correct in form but also matches physical or theoretical constraints at certain relaxed points, benefiting predictions and real-world applications.