Problem 22 - Advanced Engineering Mathematics

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 is the trivial solution: \(x(t) = 0\) and \(y(t) = 0\) for all \(t\).

Step 1: Analyze the System of Differential Equations

We are given a system of two first-order differential equations: \(\frac{dx}{dt} = y - 1\) and \(\frac{dy}{dt} = -3x + 2y\). We also have initial conditions \(x(0) = 0\) and \(y(0) = 0\). The goal is to solve this system to find functions \(x(t)\) and \(y(t)\).

Step 2: Determine Eigenvalues of the System

Form the matrix \(A = \begin{pmatrix} 0 & 1 \ -3 & 2 \end{pmatrix}\) from the coefficients of \(x\) and \(y\) in the equations. The eigenvalues are found by solving \(\text{det}(A - \lambda I) = 0\). This yields the characteristic equation \(\lambda^2 - 2\lambda + 3 = 0\), which we solve to find eigenvalues \(\lambda_1 = 1 + i\sqrt{2}\) and \(\lambda_2 = 1 - i\sqrt{2}\).

Step 3: Find Eigenvectors Corresponding to Eigenvalues

For each eigenvalue, solve \((A - \lambda I)\mathbf{v} = 0\) to find the corresponding eigenvector. For \(\lambda_1 = 1 + i\sqrt{2}\), the eigenvector is \(\begin{pmatrix} 1 \ 1 + i\sqrt{2} \end{pmatrix}\). For \(\lambda_2 = 1 - i\sqrt{2}\), the eigenvector is \(\begin{pmatrix} 1 \ 1 - i\sqrt{2} \end{pmatrix}\).

Step 4: Form the General Solution of the System

The general solution is a combination of the solutions from each eigenvalue. It can be expressed as:\[\begin{pmatrix} x(t) \ y(t) \end{pmatrix} = c_1 e^{(1+i\sqrt{2})t} \begin{pmatrix} 1 \ 1 + i\sqrt{2} \end{pmatrix} + c_2 e^{(1-i\sqrt{2})t} \begin{pmatrix} 1 \ 1 - i\sqrt{2} \end{pmatrix}\]

Step 5: Apply Initial Conditions

Apply the initial conditions \(x(0) = 0\) and \(y(0) = 0\) to determine \(c_1\) and \(c_2\). Solving the equations for \(t = 0\) gives the conditions:\[c_1 \begin{pmatrix} 1 \ 1 + i\sqrt{2} \end{pmatrix} + c_2 \begin{pmatrix} 1 \ 1 - i\sqrt{2} \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\]This results in the solution \(c_1 = 0\) and \(c_2 = 0\).

Step 6: Conclude with the Solution

Since both coefficients \(c_1\) and \(c_2\) are zero, the only solution is the trivial solution \(x(t) = 0\) and \(y(t) = 0\) for all \(t\). This satisfies both the differential equations and the initial conditions.

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations encompasses both the differential equation itself and specified values of the unknown function at a given point, usually at time zero. In our exercise, we're dealing with two equations:

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

The initial conditions given as \( x(0) = 0 \) and \( y(0) = 0 \) mean that at time \( t = 0 \), both \( x \) and \( y \) start at zero. To solve an initial value problem, you integrate the equations while respecting these initial settings. The goal is to find the functions \( x(t) \) and \( y(t) \) that satisfy both the differential equations and these initial conditions, which define a unique trajectory of the system from a specified starting point.

Eigenvalues and Eigenvectors

The concepts of eigenvalues and eigenvectors play a crucial role in solving systems of linear differential equations. For our exercise, these elements provide insights into the behavior of the system over time.

Given the matrix representation of the system:\[ A = \begin{pmatrix} 0 & 1 \ -3 & 2 \end{pmatrix}\]you solve for eigenvalues \( \lambda \) via the characteristic equation \( \text{det}(A - \lambda I) = 0 \). This calculation gives eigenvalues:

• \( \lambda_1 = 1 + i \sqrt{2} \)
• \( \lambda_2 = 1 - i \sqrt{2} \)

The corresponding eigenvectors tell us the directions along which the system evolves. For \( \lambda_1 \), the eigenvector is \( \begin{pmatrix} 1 \ 1 + i\sqrt{2} \end{pmatrix} \), and for \( \lambda_2 \), it's \( \begin{pmatrix} 1 \ 1 - i\sqrt{2} \end{pmatrix} \). These vectors are indispensable for forming the general solution of the differential equations.

Linear Systems of Differential Equations

The system described falls under the category of linear systems of differential equations, mainly due to the linear nature of the equations involved. Linear systems are characterized by variables and their derivatives being of first degree and without products or higher-degree components of those variables.

The equations \( \frac{dx}{dt} = y - 1 \) and \( \frac{dy}{dt} = -3x + 2y \) form a linear system, represented by a matrix equation \( \mathbf{X}' = A\mathbf{X} + \mathbf{B} \) where \( \mathbf{X} \) is the vector of the functions \( x \) and \( y \), and \( A \) is the coefficient matrix we previously defined. Solving linear systems often involves diagonalizing the matrix \( A \) using eigenvalues and eigenvectors, allowing the system to be expressed in terms where each equation is easier to handle individually. The solution often comprises an exponential expression involving the eigenvalues, combined into a general solution for the variables. In this problem scenario, however, initial conditions led to a simple outcome, showcasing the elegance of linear systems when specific conditions provide direct solutions.