Problem 38 - Advanced Engineering Mathematics

Question

In Problems 35-46, find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=4 x+5 y \\ &\frac{d y}{d t}=-2 x+6 y \end{aligned} $$

Step-by-Step Solution

Answer

The general solution involves complex exponentials: \(\mathbf{X}(t) = e^{5t}\cos(3t) + e^{5t}i\sin(3t)\).

Step 1: Write the System in Matrix Form

The system of differential equations can be written in matrix form as \( \mathbf{X}' = A\mathbf{X} \), where \( \mathbf{X} = \begin{pmatrix} x \ y \end{pmatrix} \) and the matrix \( A = \begin{pmatrix} 4 & 5 \ -2 & 6 \end{pmatrix} \). Thus, the system is: \[ \frac{d}{dt} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 4 & 5 \ -2 & 6 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}. \]

Step 2: Find the Eigenvalues of the Coefficient Matrix

To find the eigenvalues of matrix \( A \), solve the characteristic equation: \( \det(A - \lambda I) = 0 \). Compute: \[ \det \begin{pmatrix} 4-\lambda & 5 \ -2 & 6-\lambda \end{pmatrix} = (4-\lambda)(6-\lambda) - (5)(-2). \] Simplify the determinant: \( (4-\lambda)(6-\lambda) + 10 = 0 \).

Step 3: Solve the Characteristic Equation

Expand and simplify the determinant to find the characteristic polynomial: \( \lambda^2 - 10\lambda + 34 = 0 \). Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=-10, c=34 \). Calculate the discriminant \( b^2 - 4ac = 100 - 136 = -36 \). The eigenvalues are \( \lambda_1, \lambda_2 = 5 \pm i\sqrt{9} = 5 \pm 3i \).

Step 4: Find the Corresponding Eigenvectors

To find the eigenvectors for \( \lambda = 5 + 3i \), substitute \( \lambda \) into \( (A - \lambda I) \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). This simplifies to: \( \begin{pmatrix} -1-3i & 5 \ -2 & 1-3i \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). Solve for \( v_1 \) and \( v_2 \) to find the eigenvector. Repeat for \( \lambda = 5 - 3i \).

Step 5: Write the General Solution

The general solution is a linear combination of complex exponentials: \( \mathbf{X}(t) = c_1 \begin{pmatrix} e^{(5 + 3i)t} \ e^{(5 + 3i)t} \end{pmatrix} + c_2 \begin{pmatrix} e^{(5 - 3i)t} \ e^{(5 - 3i)t} \end{pmatrix} \). Express using real parts: \[ \mathbf{X}(t) = e^{5t} (c_1 \begin{pmatrix} \cos(3t) + i\sin(3t) \ \cos(3t) + i\sin(3t) \end{pmatrix} + c_2 \begin{pmatrix} \cos(3t) - i\sin(3t) \ \cos(3t) - i\sin(3t) \end{pmatrix}). \] Simplify using Euler's formula.

Key Concepts

Matrix Form of Differential Equations

When dealing with systems of differential equations, converting them to matrix form can greatly simplify the analysis and solution process. In this particular exercise, we are given:
\[ \frac{dx}{dt} = 4x + 5y \]\[ \frac{dy}{dt} = -2x + 6y \]
These equations describe the rate of change of two variables, \(x\) and \(y\), in terms of each other. To write them in matrix form, we define a vector \( \mathbf{X} \) as:

• \( \mathbf{X} = \begin{pmatrix} x \ y \end{pmatrix} \)

Next, we express the system as:
\[ \frac{d}{dt} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 4 & 5 \ -2 & 6 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \]
Here, the matrix \( A = \begin{pmatrix} 4 & 5 \ -2 & 6 \end{pmatrix} \) is the coefficient matrix. Each element of this matrix specifies how the variables influence each other's rates of change. This matrix formulation allows us to apply powerful algebraic techniques to solve the system.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are central concepts in solving matrix differential equations. They provide insights into the behavior and solutions of the system.

• To find eigenvalues, we solve the characteristic equation: \( \det(A - \lambda I) = 0 \).

For our matrix \( A \), the characteristic equation comes out as:
\[ \det \begin{pmatrix} 4-\lambda & 5 \ -2 & 6-\lambda \end{pmatrix} = (4-\lambda)(6-\lambda) + 10 = 0 \]
Solving this quadratic equation gives the eigenvalues \( \lambda_1 = 5 + 3i \) and \( \lambda_2 = 5 - 3i \). These complex eigenvalues indicate oscillatory behavior in the system.

• Now, to find the eigenvectors associated with each eigenvalue, we solve \( (A - \lambda I) \mathbf{v} = \mathbf{0} \).

This yields a set of equations that help us find the direction of the eigenvector. Eigenvectors indicate stable or unstable directions in the system, guiding the solution trajectory.

Complex Exponentials

Complex exponentials arise naturally as solutions to systems with complex eigenvalues. They can be expressed using Euler's formula and are crucial in understanding oscillatory systems.

• A key principle is Euler's formula: \( e^{i \theta} = \cos(\theta) + i\sin(\theta) \).

In this exercise, the eigenvalues were complex, leading to solutions in terms of complex exponentials:
\[ e^{(5 \pm 3i)t} \]
These correspond to sinusoidal functions combined with exponential growth or decay, as:
\[ e^{5t}(\cos(3t) + i\sin(3t)) \]
Thus, the general solution combines these forms, expressed in terms of real coefficients:
\[ \mathbf{X}(t) = c_1 e^{(5 + 3i)t} + c_2 e^{(5 - 3i)t} \]
By using Euler's formula, we simplify the complex exponentials into sines and cosines, highlighting the oscillatory part of the solution. This step helps in interpreting the physical or geometrical behavior of the system.