Problem 11 - Advanced Engineering Mathematics

Question

Solve each differential equation by variation of parameters. $$ y^{\prime \prime}+3 y^{\prime}+2 y=\frac{1}{1+e^{x}} $$

Step-by-Step Solution

Answer

The general solution is: \( y = C_1 e^{-x} + C_2 e^{-2x} + x e^{-x} - \ln|1+e^x| e^{-x} - \ln|1+e^x| e^{-2x} \).

Step 1: Find the Complementary Function

First, solve the homogeneous equation: \[ y'' + 3y' + 2y = 0 \]The characteristic equation is:\[ r^2 + 3r + 2 = 0 \]Factoring gives you: \[ (r + 1)(r + 2) = 0 \]So, the roots are \(-1\) and \(-2\). Therefore, the complementary function is:\[ y_c = C_1 e^{-x} + C_2 e^{-2x} \]

Step 2: Choose a Particular Solution Form

Given that the solution involves variation of parameters, let's use the complementary solution \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) to find a particular solution \( y_p = u_1 y_1 + u_2 y_2 \), where \( u_1 \) and \( u_2 \) are functions to be determined.

Step 3: Setup the Variation Parameters Equations

The Wronskian \( W \) of \( y_1 \) and \( y_2 \) is given by:\[ W = \begin{vmatrix} e^{-x} & e^{-2x} \ -e^{-x} & -2e^{-2x} \end{vmatrix} = e^{-x}(-2e^{-2x}) - (-e^{-x})(e^{-2x}) = -e^{-3x} \]Use these equations to find \( u_1' \) and \( u_2' \):\[ u_1' = \frac{-y_2 g(x)}{W} = \frac{-e^{-2x} \cdot \frac{1}{1+e^x}}{-e^{-3x}} = \frac{e^x}{1+e^x} \]\[ u_2' = \frac{y_1 g(x)}{W} = \frac{e^{-x} \cdot \frac{1}{1+e^x}}{-e^{-3x}} = -\frac{1}{1+e^x} \]

Step 4: Integrate the Variation Parameters

Integrate to find \( u_1 \) and \( u_2 \):\[ u_1 = \int \frac{e^x}{1+e^x} \, dx = \int \left(1 - \frac{1}{1+e^x}\right) \, dx = x - \ln|1+e^x| + C_1 \]\[ u_2 = \int -\frac{1}{1+e^x} \, dx = -\ln|1+e^x| + C_2 \]

Step 5: Find the Particular Solution

Now, find the particular solution using:\[ y_p = u_1 y_1 + u_2 y_2 \]Substitute \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \):\[ y_p = \left( x - \ln|1+e^x| \right)e^{-x} + \left( - \ln|1+e^x| \right)e^{-2x} \]Simplify to:\[ y_p = x e^{-x} - \ln|1+e^x| \, e^{-x} - \ln|1+e^x| \, e^{-2x} \]

Step 6: Write the General Solution

The general solution is the sum of the complementary and particular solutions:\[ y = y_c + y_p = C_1 e^{-x} + C_2 e^{-2x} + x e^{-x} - \ln|1+e^x| \, e^{-x} - \ln|1+e^x| \, e^{-2x} \]

Key Concepts

differential equations

A differential equation is a mathematical equation that relates a function with its derivatives. In these equations, the function represents a physical quantity, while its derivatives show how that quantity changes. For example, the equation given in this exercise, \( y'' + 3y' + 2y = \frac{1}{1+e^x} \), involves the second derivative \( y'' \), the first derivative \( y' \), and the function itself \( y \).
Differential equations are widely used to model real-world systems such as population growth, heat conduction, and electrical circuits. There are various methods to solve differential equations, including the homogeneous solution and particular solution, leading us into the topics below.

• Homogeneous Equations: These equations do not include a separate function of the independent variable, leading to solutions that describe the system's natural behavior.
• Non-homogeneous Equations: These incorporate an additional function (like \( \frac{1}{1+e^x} \) in our example) and describe how the system behaves under external influences.

Differential equations play a crucial role in understanding complex systems by describing continuous changes over time.

complementary function

The complementary function is a solution to the homogeneous differential equation, which is derived by setting the non-homogeneous part to zero. In our example, this means solving \( y'' + 3y' + 2y = 0 \).
Solving this involves finding the characteristic equation, which is done by substituting the trial solution \( y = e^{rt} \) into the homogeneous equation. This gives us the characteristic equation \( r^2 + 3r + 2 = 0 \). By factoring it to \( (r + 1)(r + 2) = 0 \), we find the roots \(-1\) and \(-2\).
These roots represent the exponential terms in the complementary function:

• Root \(-1\): Corresponds to the term \( C_1 e^{-x} \) in the solution.
• Root \(-2\): Corresponds to the term \( C_2 e^{-2x} \) in the solution.

Thus, the complementary function for the differential equation becomes \( y_c = C_1 e^{-x} + C_2 e^{-2x} \). The constants \( C_1 \) and \( C_2 \) are determined by initial or boundary conditions of the specific problem.

particular solution

In the context of differential equations, the particular solution is a specific solution to the non-homogeneous equation. It complements the complementary function to form the general solution of the non-homogeneous equation.
For our example, we use the technique of variation of parameters to find a particular solution \( y_p \). We first express \( y_p \) as a linear combination of the complementary solutions, using functions \( u_1 \) and \( u_2 \): \[ y_p = u_1 y_1 + u_2 y_2 \] Where \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) are solutions from the complementary function.
To find \( u_1 \) and \( u_2 \), we set up and solve the system of equations involving the Wronskian (explained below) and the non-homogeneous term. Integration of these functions gives us:

• \( u_1 = x - \ln|1+e^x| + C_1 \)
• \( u_2 = - \ln|1+e^x| + C_2 \)

Substituting these into the expression for \( y_p \), we derive the particular solution. The complete solution to the differential equation is thus the sum of this particular solution and the complementary function.

Wronskian

The Wronskian is a determinant used in the analysis of solutions to differential equations. For two functions \( y_1 \) and \( y_2 \), the Wronskian is a function of the independent variable constructed as follows: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]
In our example, the Wronskian of \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) is calculated to be \( -e^{-3x} \). It helps us verify that \( y_1 \) and \( y_2 \) form a fundamental set of solutions—meaning they are linearly independent.
The Wronskian is also fundamental for the variation of parameters method. By ensuring \( y_1 \) and \( y_2 \) are independent, we can guarantee the existence of the functions \( u_1 \) and \( u_2 \) needed for the particular solution.
Key properties of the Wronskian include:

• Non-zero: A non-zero Wronskian at some point ensures linear independence of the solution functions.
• Sign: The positive or negative sign doesn't affect its ability to function in variation of parameters but should be accurately calculated for consistent results.

Understanding the Wronskian equips students with a powerful tool in solving differential equations effectively.