Problem 19 - Advanced Engineering Mathematics

Question

Solve the given differential equation by variation of parameters. $$ x y^{\prime \prime}-4 y^{\prime}=x^{4} $$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 + C_2 x^5 - \frac{x^5}{25} + \frac{1}{5} x^5 \ln |x| \).

Step 1: Rewrite the Differential Equation

First, rewrite the differential equation in standard form. The original equation is \( x y'' - 4y' = x^4 \). Dividing all terms by \( x \), we get: \[ y'' - \frac{4}{x}y' = x^3 \].

Step 2: Solve the Homogeneous Equation

Solve the corresponding homogeneous equation \( y'' - \frac{4}{x}y' = 0 \). This is an Euler-Cauchy equation, and has the characteristic equation \( r(r-1) - 4r = 0 \). Simplifying, we have \( r^2 - 5r = 0 \), leading to roots \( r_1 = 0 \) and \( r_2 = 5 \). The general solution of the homogeneous equation is \( y_h = C_1 + C_2 x^5 \).

Step 3: Find Particular Solution Using Variation of Parameters

For variation of parameters, assume a solution of the form \( y_p = u_1(x) + u_2(x)x^5 \), where \( u_1' \) and \( u_2' \) satisfy:\[ u_1' \cdot 1 + u_2' \cdot x^5 = 0 \] \[ u_1' \cdot 0 + u_2' \cdot 5x^4 = x^3 \]. From the first equation, we have \( u_1' = -u_2' x^5 \). Substitute \( u_1' = -u_2' x^5 \) in the second equation:\[ 5x^4 u_2' = x^3 \]. So, \( u_2' = \frac{1}{5x} \). Integrating, we get \( u_2 = \frac{1}{5} \ln|x| \).

Step 4: Determine Functions u(x)

Using \( u_1' = -u_2' x^5 = -\frac{x^4}{5} \), integrate to find \( u_1(x) \): \[ u_1 = \int -\frac{x^4}{5} dx = -\frac{x^5}{25} \].Thus, the particular solution is\[ y_p = u_1 + u_2 x^5 = -\frac{x^5}{25} + \frac{1}{5} x^5 \ln |x| \].

Step 5: Write General Solution

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:\[ y = y_h + y_p = C_1 + C_2 x^5 - \frac{x^5}{25} + \frac{1}{5} x^5 \ln |x| \].

Key Concepts

Differential Equations

A differential equation is an equation that involves a function and its derivatives. It shows the relationship between these elements. They come in various forms, such as ordinary differential equations (ODEs) and partial differential equations (PDEs). In our exercise, we handle an ordinary differential equation, specifically concerned with one dependent variable and its derivatives regarding one independent variable, which is what categorizes it as an ODE. Differential equations are fundamental in modeling real-world phenomena, from physics to economics. They're often used to describe systems that change over time, like population dynamics or the movement of pendulums. In this particular case, the equation of interest is given in a non-standard format, but the goal is to manipulate it into a form that allows for solution techniques like variation of parameters.

Euler-Cauchy Equation

The Euler-Cauchy equation is a type of differential equation characterized by having coefficients that are powers of the independent variable. It is also known as the equidimensional equation. This kind of equation typically appears in the form:

• \( a x^2 y'' + b x y' + c y = 0 \)

To solve the Euler-Cauchy equation, we employ substitution techniques to transform it into a simpler constant-coefficient equation, making it easier to manage. By rewriting our differential equation:

• \( y'' - \frac{4}{x} y' = 0 \)

we made it resemble an Euler-Cauchy form. We then use its characteristic equation, a polynomial whose roots guide us to the form of the solution. This involves solving for characteristic roots which in this exercise resulted into two distinctive solutions related to the roots 0 and 5.

Homogeneous Solutions

Homogeneous solutions are solutions to a differential equation set to zero (where the non-variable side of the equation is zero). In a homogeneous linear differential equation, the terms involve only the function and its derivatives. Our exercise first focuses on solving the homogeneous part of the differential equation:

• \( y'' - \frac{4}{x} y' = 0 \)

After finding the characteristic equation derived from this homogeneous equation, we find real and distinct roots. These roots help us determine the general solution to the homogeneous part, in the form:

• \( y_h = C_1 + C_2 x^5 \)

where \( C_1 \) and \( C_2 \) are constants. This general solution is a fundamental part upon which we will add the particular solution found via the variation of parameters.

Particular Solutions

A particular solution addresses the non-homogeneous part of a differential equation, the side of the equation that does not involve the function or its derivatives. In this exercise, we utilize variation of parameters, a method for finding particular solutions of differential equations, especially those which are non-homogeneous. This method stems from the principle of superposition, allowing the adjustment of constants into functions to smoothly address non-homogeneous terms. Here, we assume a solution of the form:

• \( y_p = u_1(x) + u_2(x)x^5 \)

By satisfying certain conditions, we find \( u_1 \) and \( u_2 \) functions, hence determining:

• \( y_p = -\frac{x^5}{25} + \frac{1}{5} x^5 \ln |x| \)

Thus, the complete general solution to the original differential equation comprises both the homogeneous and this particular solution.