Problem 22 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$\begin{aligned} &\frac{d x}{d t}=y-1\\\ &\frac{d y}{d t}=-3 x+2 y\\\ &x(0)=0, y(0)=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solution with initial conditions is: \( x(t) = \frac{1}{3}[e^t(\cos(\sqrt{2}t) - \sin(\sqrt{2}t))] - \frac{1}{3} \), \( y(t) = e^t(-\cos(\sqrt{2}t) - \sin(\sqrt{2}t)) \).

Step 1: Identify the Type of System

The given system of differential equations is a pair of first-order linear ordinary differential equations with constant coefficients. We can write the system in matrix form or directly solve it.

Step 2: Write the System in Matrix Form

Rewrite the given system in matrix form:\[\begin{pmatrix} \frac{dx}{dt} \\frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} 0 & 1 \-3 & 2 \end{pmatrix} \begin{pmatrix} x \y \end{pmatrix} + \begin{pmatrix} -1 \0 \end{pmatrix}.\]

Step 3: Find the Eigenvalues and Eigenvectors

Find the eigenvalues of the matrix \( A = \begin{pmatrix} 0 & 1 \ -3 & 2 \end{pmatrix} \) by solving \( \det(A - \lambda I) = 0 \). We have:\[\begin{vmatrix} 0 - \lambda & 1 \-3 & 2 - \lambda\end{vmatrix} = 0.\]The characteristic equation is \( \lambda^2 - 2\lambda + 3 = 0 \), having roots (eigenvalues) \( \lambda_1 = 1 + \sqrt{2}i \) and \( \lambda_2 = 1 - \sqrt{2}i \).

Step 4: Determine the General Solution for the Homogeneous Part

The general solution for the system based on the eigenvalues is:\[\begin{pmatrix} x(t) \y(t) \end{pmatrix} = c_1 \begin{pmatrix} \cos(\sqrt{2}t) \-3\cos(\sqrt{2}t) + \sin(\sqrt{2}t) \end{pmatrix} e^t + c_2 \begin{pmatrix} \sin(\sqrt{2}t) \-3\sin(\sqrt{2}t) - \cos(\sqrt{2}t) \end{pmatrix} e^t.\]

Step 5: Solve for the Particular Solution

To solve for the particular solution, guess a constant solution \( \begin{pmatrix} x_p \ y_p \end{pmatrix} \). Substituting into the system gives:\[-1 = y_p - 1 \quad \text{and} \quad 0 = -3x_p + 2y_p.\]Solving, we find \( y_p = 0 \) and \( x_p = -\frac{1}{3} \). The particular solution is:\[ x_p(t) = -\frac{1}{3}, \quad y_p(t) = 0. \]

Step 6: Combine General and Particular Solutions

The complete solution is the sum of the homogeneous and particular solutions:\[\begin{pmatrix} x(t) \y(t) \end{pmatrix} = c_1 \begin{pmatrix} \cos(\sqrt{2}t) \-3\cos(\sqrt{2}t) + \sin(\sqrt{2}t) \end{pmatrix} e^t + c_2 \begin{pmatrix} \sin(\sqrt{2}t) \-3\sin(\sqrt{2}t) - \cos(\sqrt{2}t) \end{pmatrix} e^t + \begin{pmatrix} -\frac{1}{3} \0 \end{pmatrix}. \]

Step 7: Apply Initial Conditions

Use the initial conditions \( x(0) = 0 \) and \( y(0) = 0 \) to solve for the constants \( c_1 \) and \( c_2 \). From \( x(0) = 0 \), solve:\[-\frac{1}{3} + c_1 = 0 \rightarrow c_1 = \frac{1}{3}.\]From \( y(0) = 0 \), solve:\[-c_1 + c_2 \times 1 = 0 \rightarrow \frac{1}{3} + c_2 \times 1 = 0.\]Thus, \( c_2 = -\frac{1}{3} \).

Step 8: Write the Final Solution

Substitute the values of \( c_1 \) and \( c_2 \) into the full solution:\[\begin{pmatrix} x(t) \y(t) \end{pmatrix} = \left( \frac{1}{3} \right)\begin{pmatrix} \cos(\sqrt{2}t) \-3\cos(\sqrt{2}t) + \sin(\sqrt{2}t) \end{pmatrix} e^t - \left( \frac{1}{3} \right)\begin{pmatrix} \sin(\sqrt{2}t) \-3\sin(\sqrt{2}t) - \cos(\sqrt{2}t) \end{pmatrix} e^t + \begin{pmatrix} -\frac{1}{3} \0 \end{pmatrix}. \]

Key Concepts

First-Order Linear Differential Equations

Understanding first-order linear differential equations is crucial for solving various mathematical and scientific problems. A first-order linear differential equation is an equation that involves the derivative of a function and the function itself, with the highest derivative being first-order.
This means, in simple terms, the equation only contains the first derivative of the function.A typical form of such an equation is:

• \( \frac{dy}{dt} + P(t) y = Q(t) \)

Here, \( y \) is the function we want to solve for, \( P(t) \) and \( Q(t) \) are known functions of \( t \).In the exercise, we've encountered two coupled first-order linear differential equations:

• \( \frac{dx}{dt} = y - 1 \)
• \( \frac{dy}{dt} = -3x + 2y \)

These equations describe how two variables, \( x \) and \( y \), change over time. They depend on each other as well as themselves. The goal is to find solutions for \( x(t) \) and \( y(t) \) given some initial conditions. This is a common form seen in dynamical systems, modeling various processes such as electrical circuits, economies, or biological systems.

Matrix Form Solution

Converting differential equations into a matrix form can simplify their analysis and solution. This method is very useful, especially when dealing with systems of equations.The matrix form of a system of linear differential equations groups these equations into a single equation:

• \[\begin{pmatrix} \frac{dx}{dt} \ \frac{dy}{dt} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} + \begin{pmatrix} b_1 \ b_2 \end{pmatrix} \]

In our case, the matrix is:

• \[\begin{pmatrix} 0 & 1 \ -3 & 2 \end{pmatrix} \]

The left side represents the derivatives, while the right side contains a coefficient matrix, which operates on the vector of the functions \( x(t) \) and \( y(t) \), along with a constant vector.Writing the system in matrix form helps in understanding the role each equation plays in the system and forms the basis for finding solutions through matrices' eigenvalues and eigenvectors. It's especially beneficial since this transformation often reveals patterns and symmetries in the system.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are powerful tools in understanding the dynamic behavior of systems of differential equations. When a system is expressed in matrix form, as seen in our exercise, solving for eigenvalues and eigenvectors provides insight into the system's stability and oscillatory behavior.To find the eigenvalues, you solve the characteristic equation:

• \[ \text{det}(A - \lambda I) = 0 \]

where \( A \) is the matrix of coefficients, and \( \lambda \) is the eigenvalue. Here, the matrix \( A \) is:

• \[\begin{pmatrix} 0 & 1 \ -3 & 2 \end{pmatrix} \]

The characteristic equation for this matrix yields complex eigenvalues \( 1 \pm \sqrt{2}i \), indicating oscillatory solutions or cycles in the system.Once eigenvalues are determined, corresponding eigenvectors can be calculated. These eigenvectors help in constructing the solution to the differential equations. In our solution, they combine with exponential terms to form the general solution. Eigenvectors essentially provide directions in which the system stretches or compresses, guiding the behavior of solutions over time.