Problem 22 - Advanced Engineering Mathematics

Question

The eigenvalues are \(\lambda_{1}=\frac{1}{4}\) and \(\lambda_{2}=\frac{1}{2} .\) This leads to the system $$\begin{array}{l} e^{t / 4}=b_{0}+\frac{1}{4} b_{1} \\ e^{t / 2}=b_{0}+\frac{1}{2} b_{1} \end{array}$$ which has the solution \(b_{0}=2 e^{t / 4}+e^{t / 2}\) and \(b_{1}=-4 e^{t / 4}+4 e^{t / 2} .\) Then $$e^{\mathbf{A} t}=b_{0} \mathbf{I}+b_{1} \mathbf{A}=\left(\begin{array}{cc} -2 e^{t / 4}+3 e^{t / 2} & 6 e^{t / 4}-6 e^{t / 2} \\ -e^{t / 4}+e^{t / 2} & 3 e^{t / 4}-2 e^{t / 2} \end{array}\right)$$ The general solution of the system is then $$\begin{aligned} \mathbf{X}=e^{\mathbf{A} t} \mathbf{C} &=\left(\begin{array}{cc} -2 e^{t / 4}+3 e^{t / 2} & 6 e^{t / 4}-6 e^{t / 2} \\ -e^{t / 4}+e^{t / 2} & 3 e^{t / 4}-2 e^{t / 2} \end{array}\right)\left(\begin{array}{c} c_{1} \\ c_{2} \end{array}\right) \\ &=c_{1}\left(\begin{array}{c} -2 \\ -1 \end{array}\right) e^{t / 4}+c_{1}\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2}+c_{2}\left(\begin{array}{c} 6 \\ 3 \end{array}\right) e^{t / 4}+c_{2}\left(\begin{array}{c} -6 \\ -2 \end{array}\right) e^{t / 2} \\ &=\left(-c_{1}+3 c_{2}\right)\left(\begin{array}{c} 2 \\ 1 \end{array}\right) e^{t / 4}+\left(c_{1}-2 c_{2}\right)\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2} \\ &=c_{3}\left(\begin{array}{c} 2 \\ 1 \end{array}\right) e^{t / 4}+c_{4}\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2}. \end{aligned}$$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X} = c_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} e^{t/4} + c_4 \begin{pmatrix} 3 \\ 1 \end{pmatrix} e^{t/2} \).

Step 1: Recognize the System of Equations

We begin with the given system of equations derived from the eigenvalues and exponentials: \( e^{t/4} = b_0 + \frac{1}{4}b_1 \) and \( e^{t/2} = b_0 + \frac{1}{2}b_1 \). These equations will help us find the coefficients \(b_0\) and \(b_1\) in terms of \(e^{t/4}\) and \(e^{t/2}\).

Step 2: Solve for Coefficients

Given solutions for \( b_0 \) and \( b_1 \): \( b_0 = 2e^{t/4} + e^{t/2} \) and \( b_1 = -4e^{t/4} + 4e^{t/2} \), these are used to construct the matrix exponential \( e^{\mathbf{A}t} \).

Step 3: Construct Matrix Exponential

Substitute \( b_0 \) and \( b_1 \) into the expression for \( e^{\mathbf{A}t} = b_0 \mathbf{I} + b_1 \mathbf{A} \) to obtain the matrix:\[egrho \begin{array}{cc}-2e^{t/4} + 3e^{t/2} & 6e^{t/4} - 6e^{t/2} \-e^{t/4} + e^{t/2} & 3e^{t/4} - 2e^{t/2} \\end{array} egrho\]

Step 4: Express the General Solution

Multiply \( e^{\mathbf{A}t} \) by a constant vector \( \mathbf{C} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} \) to form the general solution:\[egrho \mathbf{X} = e^{\mathbf{A}t} \mathbf{C} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} egrho\]This results in:\[c_1 \begin{pmatrix} -2 \ -1 \end{pmatrix} e^{t/4} + c_1 \begin{pmatrix} 3 \ 1 \end{pmatrix} e^{t/2} + c_2 \begin{pmatrix} 6 \ 3 \end{pmatrix} e^{t/4} + c_2 \begin{pmatrix} -6 \ -2 \end{pmatrix} e^{t/2} \]

Step 5: Simplify and Rearrange

Simplify the expressions by combining similar terms and factoring out coefficients:\[= \begin{pmatrix} (-c_1 + 3c_2)e^{t/4} & (c_1 - 2c_2)e^{t/2} \end{pmatrix} \begin{pmatrix} 2 \ 1 \end{pmatrix}+ \begin{pmatrix} 3 \ 1 \end{pmatrix} \]Rewrite using new constants \( c_3 = -c_1 + 3c_2 \) and \( c_4 = c_1 - 2c_2 \).

Step 6: Finalize Solution Form

The general solution simplifies to the form using newly defined constants \(c_3 \) and \(c_4 \):\[\mathbf{X} = c_3 \begin{pmatrix} 2 \ 1 \end{pmatrix} e^{t/4} + c_4 \begin{pmatrix} 3 \ 1 \end{pmatrix} e^{t/2}\]This expresses the solution in terms of \( e^{t/4} \) and \( e^{t/2} \) with vectors indicating component contributions.

Key Concepts

Eigenvalues and Eigenvectors

In mathematics, the concepts of eigenvalues and eigenvectors play a crucial role in understanding linear transformations and their applications. An eigenvalue is a scalar that indicates how much its corresponding eigenvector is stretched during a linear transformation. Meanwhile, the eigenvector remains pointing in the same direction, merely scaled by the eigenvalue. To find these, one typically solves the characteristic equation, which is derived from the matrix of the system.

For a given matrix \( \mathbf{A} \), we find values of \( \lambda \) for which the equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) holds true, where \( \mathbf{I} \) is the identity matrix. Solving this equation yields the eigenvalues. Then, for each eigenvalue, solving the equation \((\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0}\) yields the corresponding eigenvectors. These vectors represent directions of pure scaling within the transformation space.

In our exercise, the eigenvalues found are \( \lambda_1 = \frac{1}{4} \) and \( \lambda_2 = \frac{1}{2} \). These determine the system behavior and help in constructing solutions to differential equations by allowing the separation of time dynamics from spatial orientation.

System of Differential Equations

A system of differential equations involves multiple equations that describe how a set of variables changes with respect to another variable, often time. It is a powerful tool used in modeling real-life phenomena such as population growth, electrical circuits, and in our case, linear dynamic systems.

In the provided exercise, we're dealing with a linear system represented by the matrix \( \mathbf{A} \). The system of equations derived from the eigenvalues \( e^{t/4} = b_0 + \frac{1}{4}b_1 \) and \( e^{t/2} = b_0 + \frac{1}{2}b_1 \) helps in determining specific coefficients for building expressions of the matrix exponential.

Understanding the behavior of differential equations often requires techniques such as examining their stability, calculating eigenvalues, and using matrix exponentials to find solutions. These steps are key when you're trying to understand how the system evolves over time from given initial conditions.

General Solution of a System

The general solution of a system of equations provides a complete description of all possible solutions, especially important in the context of differential equations. It usually involves constants that can be adjusted for particular initial conditions.

In solving the differential system \( \mathbf{X} = e^{\mathbf{A}t} \mathbf{C} \), the matrix \( e^{\mathbf{A}t} \) is determined using coefficients \( b_0 \) and \( b_1 \), where \( b_0 = 2e^{t/4} + e^{t/2} \) and \( b_1 = -4e^{t/4} + 4e^{t/2} \). This matrix algebraically encapsulates the dynamics dictated by the eigenvalues and eigenvectors found earlier.

The general solution is expressed as a combination of these time-dependent exponential matrices multiplied by the vector \( \mathbf{C} = \begin{pmatrix} c_1 \ c_2 \end{pmatrix} \), which represents arbitrary constants from initial conditions. In simplified form, it results in linear combinations like \( c_3 \begin{pmatrix} 2 \ 1 \end{pmatrix} e^{t/4} + c_4 \begin{pmatrix} 3 \ 1 \end{pmatrix} e^{t/2} \), succinctly showing how solutions evolve across time through these constants.