Problem 12 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Find the Complementary Solution

First, use the auxiliary equation to solve the homogeneous equation. The given differential equation is \( y'' - 2y' + y = 0 \). The characteristic equation is \( r^2 - 2r + 1 = 0 \), which factors to \((r-1)^2 = 0\). The root is \( r = 1 \) with multiplicity 2. Therefore, the complementary solution is \( y_c = C_1 e^x + C_2 x e^x \).

Step 2: Setup for Variation of Parameters

Identify \( y_1 = e^x \) and \( y_2 = xe^x \) from the complementary solution \( y_c = C_1 y_1 + C_2 y_2 \). The particular solution using variation of parameters requires finding functions \( u_1(x) \) and \( u_2(x) \).

Step 3: Calculate Wronskian

Calculate the Wronskian of \( y_1 \) and \( y_2 \). The Wronskian \( W(y_1, y_2) \) is computed as follows: \[ W = \begin{vmatrix} e^x & xe^x \ e^x & (x+1)e^x \end{vmatrix} = e^{2x} \].

Step 4: Find Functions \( u_1(x) \) and \( u_2(x) \)

The formulas for \( u_1' \) and \( u_2' \) are given by: \[ u_1' = \frac{-y_2 \cdot g(x)}{W} \quad \text{and} \quad u_2' = \frac{y_1 \cdot g(x)}{W} \] where \( g(x) = \frac{e^x}{1+x^2} \).Calculate \( u_1' = -\frac{xe^x \cdot \frac{e^x}{1+x^2}}{e^{2x}} = -\frac{x}{1+x^2} \) and \( u_2' = \frac{e^x \cdot \frac{e^x}{1+x^2}}{e^{2x}} = \frac{1}{1+x^2} \).

Step 5: Integrate to Find \( u_1(x) \) and \( u_2(x) \)

Integrate \( u_1' \) and \( u_2' \):\[ u_1(x) = \int -\frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C_3 \]\[ u_2(x) = \int \frac{1}{1+x^2} \, dx = \arctan(x) + C_4 \].

Step 6: Write the Particular Solution

The particular solution is \( y_p = u_1(x) y_1 + u_2(x) y_2 \). Substitute the values: \[ y_p = \left( \frac{1}{2} \ln(1+x^2) \right) e^x + \left( \arctan(x) \right) x e^x \].

Step 7: Write the General Solution

Combine the complementary and particular solutions to write the general solution:\[ y = C_1 e^x + C_2 x e^x + \frac{1}{2} \ln(1+x^2) e^x + \arctan(x) x e^x \].

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They appear frequently in different areas of science and engineering to model how quantities change over time or space.
In the context of this problem, we are dealing with a second-order linear differential equation of the form:

• Homogeneous Part: This part is represented by the equation without any external function, i.e., \( y'' - 2y' + y = 0 \).
• Non-Homogeneous Part: This part includes an external function, represented as \( \frac{e^x}{1+x^2} \).

The general solution for such equations is often a combination of two parts: the complementary solution (from the homogeneous part) and the particular solution (from the non-homogeneous part). Understanding each component is crucial to finding the overall solution.

Complementary Solution

The complementary solution, denoted as \( y_c \), arises from solving the homogeneous part of a differential equation. For the given problem, we start with the equation \( y'' - 2y' + y = 0 \).
To find \( y_c \), we form the characteristic equation:

• Characteristic equation: \( r^2 - 2r + 1 = 0 \).
• This equation factors to \((r-1)^2 = 0 \), giving a repeated root \( r = 1 \).

For a repeated root such as this, the complementary solution is given by:

• \( y_c = C_1 e^x + C_2 x e^x \)

where \( C_1 \) and \( C_2 \) are constants. This form uses the exponential function \( e^x \), which is a common theme in solving linear differential equations due to its unique properties.

Wronskian

The Wronskian is a determinant used to determine whether a set of functions are linearly independent. In the method of variation of parameters, it plays a key role in finding the particular solution.
For this problem, we need the Wronskian of \( y_1 = e^x \) and \( y_2 = x e^x \), which are part of our complementary solution:

• Calculate the Wronskian: \[ W = \begin{vmatrix} e^x & xe^x \ e^x & (x+1)e^x \end{vmatrix} = e^{2x} \]

Having a non-zero Wronskian, like \( e^{2x} \) in our case, confirms the linear independence of these solutions. This ensures that we can use these functions for constructing our particular solution through variation of parameters.

Particular Solution

A particular solution accounts for the non-homogeneous part of a differential equation. In this method, we use variation of parameters which involves finding two functions, \( u_1(x) \) and \( u_2(x) \), that multiply the solutions of the complementary solution.
The formulas for these functions involve the non-homogeneous term \( g(x) = \frac{e^x}{1+x^2} \), leading to:

• \( u_1' = -\frac{x}{1+x^2} \)
• \( u_2' = \frac{1}{1+x^2} \)

By integrating \( u_1' \) and \( u_2' \), we find:

• \( u_1(x) = \frac{1}{2} \ln(1+x^2) + C_3 \)
• \( u_2(x) = \arctan(x) + C_4 \)

Finally, the particular solution \( y_p \) becomes:\[ y_p = \left( \frac{1}{2} \ln(1+x^2) \right) e^x + \left( \arctan(x) \right) x e^x \]This enables combining with the complementary solution to find the general solution.