Problem 14 - Advanced Engineering Mathematics

Question

Use variation of parameters to solve the given system. \(\frac{d x}{d t}=2 x-y\) \(\frac{d y}{d t}=3 x-2 y+4 t\)

Step-by-Step Solution

Answer

Using variation of parameters, the solution is obtained by combining the homogeneous solution and the particular solution related to the non-homogeneous term.

Step 1: Find the Homogeneous Solution

First, consider the associated homogeneous system:\[ \frac{d x}{d t} = 2x - y \]\[ \frac{d y}{d t} = 3x - 2y \]To solve this system, assume a solution of the form \( \mathbf{v} = \begin{pmatrix} x_h \ y_h \end{pmatrix} = \begin{pmatrix} e^{\lambda t} \ e^{\lambda t} \end{pmatrix} \). Substitute into the homogeneous equations to obtain the characteristic equation. After solving, determine the eigenvalues and eigenvectors to construct the general solution of the homogeneous system.

Step 2: Solve the Characteristic Equation

The characteristic equation is derived from the matrix form:\[ \begin{pmatrix} 2-\lambda & -1 \ 3 & -2-\lambda \end{pmatrix} \begin{pmatrix} x_h \ y_h \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]The determinant of this matrix \((2-\lambda)(-2-\lambda) - (-1)(3) = \lambda^2 + 1\), gives the characteristic equation: \( \lambda^2 + 1 = 0 \).Solve for \( \lambda \), finding \( \lambda = i \) and \( \lambda = -i \).

Step 3: Find Eigenvectors and Construct the Homogeneous Solution

Using \( \lambda = i \), solve \( \begin{pmatrix} 2-i & -1 \ 3 & -2-i \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix} = 0 \) to find eigenvectors. Repeat for \( \lambda = -i \). For simplification, the general solution of the homogeneous system, considering trigonometric representation, is:\[ \mathbf{v}_{h}(t) = c_1 \begin{pmatrix} \cos(t) \ 0.6\cos(t) - 0.8\sin(t) \end{pmatrix} + c_2 \begin{pmatrix} \sin(t) \ 0.8\cos(t) + 0.6\sin(t) \end{pmatrix} \] where \(c_1\) and \(c_2\) are constants.

Step 4: Set Up Variation of Parameters

The particular solution \( \mathbf{v}_{p}(t) \) is found using variation of parameters:\[ \mathbf{v}_{p}(t) = u_1(t) \mathbf{v}_{1}(t) + u_2(t) \mathbf{v}_{2}(t) \]where \( u_1(t) \) and \( u_2(t) \) are functions determined by the method. The matrix equation:\[ W(t) \begin{pmatrix} \frac{d u_1}{dt} \ \frac{d u_2}{dt} \end{pmatrix} = \mathbf{g}(t) \]where \( W(t) \) is the Wronskian matrix formed from the homogeneous solutions \( \mathbf{v}_{1}(t) \) and \( \mathbf{v}_{2}(t) \), and \( \mathbf{g}(t) = \begin{pmatrix} 0 \ 4t \end{pmatrix} \).

Step 5: Calculate the Wronskian and Solve for Particular Solution

Calculate the Wronskian:\[ W(t) = \begin{vmatrix} \cos(t) & \sin(t) \ 0.6\cos(t) - 0.8\sin(t) & 0.8\cos(t) + 0.6\sin(t) \end{vmatrix} = 1 \]Solve:\[ \begin{pmatrix} \frac{d u_1}{d t} \ \frac{d u_2}{d t} \end{pmatrix} = W(t)^{-1} \mathbf{g}(t) \ = \begin{pmatrix} 0 \ 4t \end{pmatrix} \]Solve these integrals to find \( u_1(t) \) and \( u_2(t) \), giving solutions like \( u_1(t) = \int(0) \, dt = C_1 \), and \( u_2(t) = \int(4t) \, dt = 2t^2 + C_2 \).

Step 6: Form the General Solution

Construct the general solution:\[ \mathbf{v}(t) = \mathbf{v}_h(t) + \mathbf{v}_p(t) \]Substitute the solutions for \( u_1(t) \) and \( u_2(t) \) into \( \mathbf{v}_{p}(t) \) and add to the homogeneous solution:\[ \mathbf{v}(t) = \begin{pmatrix} c_1 \cos(t) + c_2 \sin(t) \+ 2t^2 \sin(t) \ c_1 (0.6\cos(t) - 0.8\sin(t)) + c_2 (0.8\cos(t) + 0.6\sin(t)) + 2t^2 (0.8\cos(t) + 0.6\sin(t)) \end{pmatrix} \]

Key Concepts

System of Differential Equations

A system of differential equations consists of multiple equations involving derivatives of various functions. Each function represents a different aspect of the same system, all depending on the same variable, often time.
In the given exercise, we have two differential equations:

• \( \frac{d x}{d t} = 2x - y \)
• \( \frac{d y}{d t} = 3x - 2y + 4t \)

These equations work together, describing how both \( x \) and \( y \) change over time \( t \). To solve such systems, specific methods like 'variation of parameters' are applied. Here, it's crucial to first consider the associated homogeneous system, which simplifies understanding the general behavior of the solution.

Homogeneous Solution

The homogeneous solution is the general solution to a system of differential equations where the non-homogeneous parts (constants or functions added on the right-hand side) are set to zero.
In context, the homogeneous version of the system drops the term \( 4t \) from the second equation:

• \( \frac{d x}{d t} = 2x - y \)
• \( \frac{d y}{d t} = 3x - 2y \)

We assume solutions of the form \( \mathbf{v} = e^{\lambda t} \begin{pmatrix} x_h \ y_h \end{pmatrix} \) and substitute back into the system. Solving these helps in finding characteristic values, leading to the homogeneous solution. The solution \( \mathbf{v}_{h}(t) \) represents the natural dynamics of the system without any external forcing functions like \( 4t \).

Characteristic Equation

When solving the homogeneous system, we derive the characteristic equation from the system matrix. This is done by substituting the exponential form \( e^{\lambda t} \) in the assumed solutions into the system.
For the given system:

• System matrix: \( \begin{pmatrix} 2 \! - \! \lambda & -1 \ 3 & -2 \! - \! \lambda \end{pmatrix} \)

The determinant of this matrix defines the characteristic equation. For this exercise, it simplifies to \( \lambda^2 + 1 = 0 \). Solving this quadratic equation yields the characteristic roots \( \lambda = i \) and \( \lambda = -i \). These roots help in constructing the basis functions of the homogeneous solution which then become part of the general solution.

Wronskian

The Wronskian is a determinant used to analyze the linear independence of a set of solutions to a differential equation. For systems, it involves the solutions at a particular instant, usually represented as a matrix.
In the exercise's context, the Wronskian \( W(t) \) is calculated from the homogeneous solutions:

• \( W(t) = \begin{vmatrix} \cos(t) & \sin(t) \ 0.6\cos(t) - 0.8\sin(t) & 0.8\cos(t) + 0.6\sin(t) \end{vmatrix} \)
• Resulting in \( W(t) = 1 \), indicating the solutions are linearly independent.

This allows us to use these functions in the method of variation of parameters to find a particular solution to the non-homogeneous system. The Wronskian's role is crucial in determining the invertibility of the system for solving the resultant set of equations and integrating them to find the complete solution.