Problem 13 - Advanced Engineering Mathematics

Question

In Problems 13-32, use vaniation of parameters to solve the given system. $$ \begin{aligned} &\frac{d x}{d t}=3 x-3 y+4 \\ &\frac{d y}{d t}=2 x-2 y-1 \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is \(x(t) = (c_1 + e^{-t})\begin{bmatrix}1\\1\end{bmatrix} + c_2 e^{t}\begin{bmatrix}1\\0\end{bmatrix}\).

Step 1: Solve the Homogeneous System

Consider the system without the constant terms: \(\frac{dx}{dt} = 3x - 3y\) and \(\frac{dy}{dt} = 2x - 2y\). Assume solutions of the form \( x_h(t) = ce^{\lambda t}\) and \( y_h(t) = de^{\lambda t}\). Substitute into the homogeneous equation to find the eigenvalues.

Step 2: Find the Eigenvalues and Eigenvectors

Setting up the eigenvalue problem leads to the characteristic equation:\[\begin{vmatrix}3-\lambda & -3 \ 2 & -2-\lambda\end{vmatrix} = 0\]Solve this to find \(\lambda_1 = 0\) (with an eigenvector \(\begin{bmatrix}1\1\end{bmatrix}\)) and \(\lambda_2 = 1\) (with an eigenvector \(\begin{bmatrix}1\0\end{bmatrix}\)).

Step 3: Construct the Homogeneous Solution

The general solution to the homogeneous system is:\[x_h(t) = c_1 e^{0t} \begin{bmatrix}1\1\end{bmatrix} + c_2 e^{t} \begin{bmatrix}1\0\end{bmatrix}\]\[x_h(t) = c_1 \begin{bmatrix}1\1\end{bmatrix} + c_2 e^{t} \begin{bmatrix}1\0\end{bmatrix}\]

Step 4: Form the Particular Solution Using Variation of Parameters

Assume a particular solution of the form \(x_p(t) = u_1(t)\begin{bmatrix}1\1\end{bmatrix} + u_2(t) e^{t}\begin{bmatrix}1\0\end{bmatrix}\). Take derivatives and substitute back into the non-homogeneous system. This gives us a system of equations to solve for \(u_1(t)\) and \(u_2(t)\).

Step 5: Solve for \(u_1(t)\) and \(u_2(t)\)

Using the system obtained from substitution, integrate these equations:\[u_1'(t) = \frac{-3}{2}e^{-t}, \quad u_2'(t) = -\frac{3}{2}e^{-2t}\]By integrating, we get\[u_1(t) = e^{-t} + C_1, \quad u_2(t) = \frac{3}{2}e^{-t} + C_2\].

Step 6: Express the General Solution

Combine the homogeneous and particular solutions:\[x(t) = \left( C_1 \begin{bmatrix}1\1\end{bmatrix} + C_2 e^{t} \begin{bmatrix}1\0\end{bmatrix} \right) + \left( e^{-t} (\begin{bmatrix}1\1\end{bmatrix}) + \frac{3}{2}e^{-t}e^{t} (\begin{bmatrix}1\0\end{bmatrix}) \right)\].

Step 7: Simplify the General Solution

Simplifying the expressions, we get:\[x(t) = (c_1 + e^{-t})\begin{bmatrix}1\1\end{bmatrix} + c_2 e^{t} \begin{bmatrix}1\0\end{bmatrix} \]Combining and factoring where possible.

Key Concepts

System of Differential Equations

A system of differential equations involves multiple differential equations that describe the dynamics of several interrelated functions. In our case, we have two functions, \( x(t) \) and \( y(t) \), both depending on the independent variable \( t \). The system given by:

• \( \frac{dx}{dt} = 3x - 3y + 4 \)
• \( \frac{dy}{dt} = 2x - 2y - 1 \)

shows how each function's rate of change not only depends on itself but also on the other function. Solving such systems typically involves techniques that consider these interdependencies, like eigenvalues and eigenvectors, which will be discussed below.
Systems of differential equations are widely used in real-world scenarios, such as modeling populations in biology, economic systems, or even electrical circuits. Understanding the interaction between variables is key to finding meaningful solutions.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a crucial role in solving systems of differential equations, especially linear ones. These mathematical tools help simplify the system by transforming it into a form where solutions are easier to find.
To find the eigenvalues and eigenvectors, we construct a matrix from the coefficients of our homogeneous system (the system without constant terms):

• \( 3x - 3y \)
• \( 2x - 2y \)

We form a characteristic equation using determinants to find \( \lambda \), the eigenvalues. For our example, solving the determinant
\(\begin{vmatrix}3-\lambda & -3 \ 2 & -2-\lambda\end{vmatrix} = 0\)
gives the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = 1 \).
Associated with each eigenvalue is an eigenvector, which represents a direction in which the system doesn't change direction. Here, the eigenvectors are \( \begin{bmatrix}1 \ 1\end{bmatrix} \) and \( \begin{bmatrix}1 \ 0\end{bmatrix} \). By combining eigenvalues and eigenvectors, we construct the general solution to the homogeneous part of the system.

Particular Solution

Finding a particular solution involves accounting for the non-homogeneous part of the differential equation system. Here, variation of parameters is utilized, a technique that modifies the homogeneous solution to incorporate external forces or conditions present in the original equation.
We assume a form for the particular solution that includes unknown functions, say \( u_1(t) \) and \( u_2(t) \), because these functions can adjust depending on the non-homogeneous terms (like the constants 4 and -1 in our problem). The assumed form is:

• \( x_p(t) = u_1(t) \begin{bmatrix}1 \ 1\end{bmatrix} + u_2(t) e^{t} \begin{bmatrix}1 \ 0\end{bmatrix} \)

We then differentiate and substitute back into the system of equations to find expressions for the derivatives of \( u_1(t) \) and \( u_2(t) \). Solving these differential equations by integration gives us the particular solution. This method ensures the solution closely matches the behavior dictated by the non-homogeneous terms.

Homogeneous Solution

The homogeneous solution refers to solving the system without considering the constant (non-homogeneous) terms. It gives us the behavior of the system based solely on its inherent dynamics.
In our exercise, once the eigenvalues and eigenvectors are known, the homogeneous solution can be constructed. Using the eigenvectors \( \begin{bmatrix}1 \ 1\end{bmatrix} \) and \( \begin{bmatrix}1 \ 0\end{bmatrix} \), and their corresponding eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = 1 \), the general form of the homogeneous solution is:

• \( x_h(t) = c_1 e^{0t} \begin{bmatrix}1 \ 1\end{bmatrix} + c_2 e^{t} \begin{bmatrix}1 \ 0\end{bmatrix} \)

Since \( e^{0t} = 1 \), this simplifies to:

• \( x_h(t) = c_1 \begin{bmatrix}1 \ 1\end{bmatrix} + c_2 e^{t} \begin{bmatrix}1 \ 0\end{bmatrix} \)

These components form the base structure of the solution, which can be further adjusted to accommodate any external conditions presented by the non-homogeneous part, ultimately assisting in forming the complete solution.