Problem 3 - Advanced Engineering Mathematics

Question

Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=-4 x+2 y \\ &\frac{d y}{d t}=-\frac{5}{2} x+2 y \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X}(t) = c_1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} e^{-6t} \).

Step 1: Write the system in matrix form

The given system of differential equations can be represented in the matrix form \( \frac{d\mathbf{X}}{dt} = A\mathbf{X} \), where \( \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix} \) and matrix \( A \) is \( \begin{bmatrix} -4 & 2 \ -\frac{5}{2} & 2 \end{bmatrix} \).

Step 2: Find the eigenvalues of the matrix

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For matrix \( A \), the characteristic polynomial is \( \begin{vmatrix}-4-\lambda & 2 \ -\frac{5}{2} & 2-\lambda \end{vmatrix} \). Calculate the determinant, which results in \( \lambda^2 + 2\lambda - 12 = 0 \).

Step 3: Solve the characteristic polynomial

The characteristic polynomial is \( \lambda^2 + 2\lambda - 12 = 0 \). Solving this quadratic equation using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a=1, b=2, c=-12 \), results in \( \lambda = 2 \) and \( \lambda = -6 \). These are the eigenvalues of the matrix \( A \).

Step 4: Find the eigenvectors for each eigenvalue

For \( \lambda = 2 \), solve \((A - 2I)\mathbf{v} = 0\) which becomes \( \begin{bmatrix} -6 & 2 \ -\frac{5}{2} & 0 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). A possible eigenvector is \( \begin{bmatrix} 1 \ 3 \end{bmatrix} \). For \( \lambda = -6 \), solve \( (A + 6I)\mathbf{v} = 0 \) which leads to \( \begin{bmatrix} 2 & 2 \ -\frac{5}{2} & 8 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). A corresponding eigenvector is \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Step 5: Construct the general solution

The general solution is a linear combination of the eigenvectors multiplied by exponential terms involving their eigenvalues. It is expressed as: \[ \mathbf{X}(t) = c_1 \begin{bmatrix} 1 \ 3 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 1 \ -1 \end{bmatrix} e^{-6t} \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts when dealing with linear differential equations, especially in the study of systems of differential equations. These concepts help us understand the behavior of the system over time.

• **Eigenvalues (\( \oldsymbol{\lambda} \)):** These are special numbers calculated from a square matrix that provide crucial insights about the matrix. They tell us about the stability and type of solutions we can expect for our differential equations.


• **Eigenvectors:** Once eigenvalues are found, eigenvectors are non-zero vectors that alter direction by the corresponding eigenvalue's magnitude when the matrix is applied to them. They help form the basis for solutions in the vector space.
  

For example, in our system, we found eigenvalues 2 and \( -6 \), allowing us to derive specific vectors along which the solutions lie: \(\begin{bmatrix} 1 \ 3 \end{bmatrix}, \begin{bmatrix} 1 \ -1 \end{bmatrix}\).These vectors are inherently linked to the growth or decay exponential rates described by the eigenvalues.

Matrix Form of System of Equations

To analyze and solve systems of differential equations, we often recast them in matrix form. This simplifies computations and utilizes linear algebra techniques to address the problem.

• **Matrix Representation:** Convert the set of differential equations into a compact matrix format—useful for complex systems with multiple equations. Each equation's variables and coefficients align into matrices for efficient processing.


• **Consistency with Notations:** The matrix \( A \) captures interactions between variables in its rows and columns. Each entry of the matrix corresponds to terms in the equations. The solution \(\frac{d\mathbf{X}}{dt} = A\mathbf{X}\)
becomes essential in uncovering the dynamics of the system.

In our exercise, the system transformed into the matrix form \( A = \begin{bmatrix} -4 & 2 \ -\frac{5}{2} & 2 \end{bmatrix}\), which reveals how the differential relationships interact with each other.

General Solution of Linear Systems

The general solution of a linear system combines solutions derived from eigenvectors and eigenvalues to describe changes over time comprehensively.

• **Linear Combination:** This is the process of finding the general solution by summing the eigenvalue-eigenvector pairs. Each pair is multiplied by an exponential function related to its eigenvalue and constant coefficients (\( c_1,c_2 \)).


• **Time Dynamics:** The solution includes exponential terms \( e^{\lambda t} \)blended with eigenvectors guiding the solution along the eigenvalue’s direction. Initial conditions set these coefficients, tailoring the solution to specific scenarios.


• **Practical Uses:** This method applies across disciplines where the evolution of a variable system depends on such linear matrices.

In this system, the constructed general solution was \[\mathbf{X}(t) = c_1 \begin{bmatrix} 1 \ 3 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 1 \ -1 \end{bmatrix} e^{-6t}\], offering a formula capturing the system's behavior over time.