Problem 23 - Advanced Engineering Mathematics

Question

Solve the given differential equation by variation of parameters. $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=\ln x $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1 x + \frac{C_2}{x} + u_1(x) x + u_2(x) \frac{1}{x}\), where \(u_1\) and \(u_2\) are functions obtained through integration.

Step 1: Identify the Type of Equation

We recognize that the equation \(x^2 y'' + x y' - y = \ln x\) is a second-order linear non-homogeneous differential equation.

Step 2: Solve the Homogeneous Equation

Consider the associated homogeneous equation: \(x^2 y'' + x y' - y = 0\). Transform this into a Cauchy-Euler form, yielding \(m^2 + 0m - 1 = 0\). Solving, we find roots \(m = 1, -1\). Thus, the general solution to the homogeneous equation is \(y_h(x) = C_1 x + \frac{C_2}{x}\).

Step 3: Use Variation of Parameters

To solve the non-homogeneous equation, we apply the method of variation of parameters. Assume a particular solution of form \(y_p = u_1(x) x + u_2(x) \frac{1}{x}\), where \(u_1\) and \(u_2\) are functions to be determined.

Step 4: Find Equations for u_1 and u_2

Using variation of parameters, we set up the system: 1. \(u_1'(x) x + u_2'(x) \frac{1}{x} = 0\) 2. \(u_1'(x) x^2 + u_2'(x) = \ln x\)These two equations are derived by ensuring there are no second derivatives in the expression for the particular solution when substituted back into the original equation.

Step 5: Solve for u_1' and u_2'

From the first equation \(u_1'(x) x + u_2'(x) \frac{1}{x} = 0\), solve for \(u_2'\): \(u_2' = -x^2 u_1'\). Substitute this into the second equation to get \(u_1'(x) x^2 - x^2 u_1'(x) = \ln x\). Hence, \(u_1'(x) = \frac{-\ln x}{x^2}\).

Step 6: Integrate to Find u_1 and u_2

Integrate \(u_1'(x) = \frac{-\ln x}{x^2}\) to get \(u_1(x) = \int \frac{-\ln x}{x^2} dx\). Use integration by parts to solve this integral. After determining \(u_1(x)\), use \(u_2'(x) = -x^2 u_1'(x)\) to find \(u_2(x)\) by integrating \(u_2'(x)\). This will result in the functions necessary to find the particular solution.

Step 7: Write the General Solution

The general solution is given by combining the homogeneous and particular solutions: \[ y(x) = C_1 x + \frac{C_2}{x} + u_1(x) x + u_2(x) \frac{1}{x} \]Substitute the expressions for \(u_1\) and \(u_2\) obtained in Step 6 into this formula to get the final solution.

Key Concepts

Second-order Linear Differential Equation

A second-order linear differential equation is a type of differential equation that involves the second derivative of a function. These equations take the general form: \[ a(x) y'' + b(x) y' + c(x) y = f(x) \] where \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the function itself.

• \( a(x), b(x), \) and \( c(x) \) are functions of \( x \)
• \( f(x) \) is a known function, making the equation non-homogeneous if \( f(x) eq 0 \).

These equations are central in mathematical modeling, engineering, and physics, as they can describe a variety of physical phenomena.
In the given exercise, our equation is \( x^2 y'' + x y' - y = \ln x \), which is non-homogeneous due to the presence of \( \ln x \).
The solution of such equations typically involves finding both the homogeneous solution (where \( f(x) = 0 \)) and a particular solution for the entire equation.

Cauchy-Euler Equation

The Cauchy-Euler equation is a specific form of a linear differential equation that can be recognized by its structure in terms of polynomials of the main variable: \[ x^2 y'' + a x y' + b y = 0 \] where \( a \) and \( b \) are constants. It is particularly useful because its solutions often involve powers of \( x \).

• Solutions take the form \( y = x^m \), where \( m \) is a constant.
• This leads to the characteristic equation \( m^2 + (a-1)m + b = 0 \) after substituting back into the differential equation.

Solving this characteristic equation provides the roots needed to form the general solution.
In our step-by-step solution, the equation \( x^2 y'' + x y' - y = 0 \) was transformed into the characteristic equation \( m^2 - 1 = 0 \) with roots \( m = 1 \) and \( m = -1 \).
These roots give us the homogeneous solution \( y_h(x) = C_1 x + \frac{C_2}{x} \).

Homogeneous and Non-homogeneous Differential Equations

Differential equations can either be homogeneous or non-homogeneous. Understanding the difference between these two types helps in formulating the right approach to solve them. **Homogeneous Differential Equations** have the form: \[ a(x) y'' + b(x) y' + c(x) y = 0 \]

• These only involve the function and its derivatives without any external function on the right-hand side.
• Solutions are typically found using characteristic equations and variable substitutions.

**Non-homogeneous Differential Equations** introduce an additional function \( f(x) \) on the right-hand side: \[ a(x) y'' + b(x) y' + c(x) y = f(x) \]

• The preferred method to solve these includes finding the general solution of the homogeneous equation and a particular solution to account for \( f(x) \).
• The complete solution is a sum of the homogeneous solution and the particular solution.

In the exercise, we used the *Variation of Parameters* method, suitable for non-homogeneous equations, to solve \( x^2 y'' + x y' - y = \ln x \).
The combination of solutions and the particular solution gave us the comprehensive solution to the differential equation.