Problem 30 - Advanced Engineering Mathematics

Question

In Problems 25-30, solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}-5 x y^{\prime}+8 y=8 x^{6}, y\left(\frac{1}{2}\right)=0, y^{\prime}\left(\frac{1}{2}\right)=0 $$

Step-by-Step Solution

Answer

Find the general solution, apply initial conditions to find constants, graph the solution curve.

Step 1: Recognize the Type of Equation

The given equation is \(x^2 y'' - 5xy' + 8y = 8x^6\). This is a second-order linear ordinary differential equation with variable coefficients. Our task is to find a function \(y(x)\) that satisfies the given equation and initial conditions.

Step 2: Solve the Homogeneous Equation

The homogeneous equation is \(x^2 y'' - 5xy' + 8y = 0\). To solve it, assume a solution of the form \(y = x^m\). Substitute \(y = x^m\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the homogeneous equation and simplify to obtain the indicial equation. Solve the indicial equation for \(m\).

Step 3: Find Particular Solution Using Method of Undetermined Coefficients

For the non-homogeneous part \(8x^6\), we assume a particular solution of the form \(y_p = Ax^6\). Differentiate to find \(y_p'\) and \(y_p''\), and substitute into the original differential equation. Solve for the coefficient \(A\) by ensuring the left-hand side equals \(8x^6\).

Step 4: General Solution Combination

Combine the general solution of the homogeneous equation and the particular solution: \(y(x) = c_1 y_1(x) + c_2 y_2(x) + y_p(x)\), where \(y_1(x)\) and \(y_2(x)\) are solutions to the homogeneous equation and \(y_p(x)\) is the particular solution.

Step 5: Apply Initial Conditions

Use the initial conditions \(y(\frac{1}{2}) = 0\) and \(y'(\frac{1}{2}) = 0\) to create a system of equations. Substitute \(x = \frac{1}{2}\) into the general solution and its derivative, and solve for the constants \(c_1\) and \(c_2\).

Step 6: Solution Verification and Graphing

Verify the solution satisfies both the differential equation and initial conditions. Use a graphing utility to plot \(y(x)\). Check that the plotted solution meets the expected initial conditions and behavior.

Key Concepts

Second-order Differential Equations

A second-order differential equation involves the second derivative of a function, which indicates how the rate of change itself is changing. These types of equations are often written as \( y'' + p(x)y' + q(x)y = g(x) \) where \( y'' \) is the second derivative, \( y' \) is the first derivative, \( y \) is the original function, and \( p(x), q(x), \) and \( g(x) \) are known functions.
In the exercise, we have a second-order differential equation: \( x^2 y'' - 5xy' + 8y = 8x^6 \). This equation is classified as linear because the terms \( y, y', y'' \) are not multiplied by each other. Linear second-order differential equations can often be solved by finding the homogeneous and particular solutions, then combining them.

Variable Coefficients

When the coefficients of a differential equation are functions of the independent variable, the equation is said to have variable coefficients. In the given equation \( x^2 y'' - 5xy' + 8y = 8x^6 \), the coefficients \( x^2 \) and \( -5x \) are functions of \( x \), hence it has variable coefficients.
Handling variable coefficients can complicate the solution process compared to constant coefficients. Solutions often involve assuming a form for the solution, such as \( y = x^m \), to find particular expressions by solving the auxiliary or indicial equation. This process is crucial for determining the complete solution of the differential equation.

Initial-value Problems

An initial-value problem involves finding a solution to a differential equation that satisfies given initial conditions at a specific point. This typically involves:

• Solving the differential equation generally.
• Applying the initial conditions to find specific constants in the solution.

In the original exercise, initial conditions are given as \( y(\frac{1}{2}) = 0 \) and \( y'(\frac{1}{2}) = 0 \). By using these conditions, a system of equations is formed that allows us to solve for the constants of integration, ensuring the solution meets the specific initial scenario provided. Initial-value problems are fundamental in determining unique solutions among many possible ones.

Method of Undetermined Coefficients

This method is a strategic way to find a particular solution of a non-homogeneous linear differential equation. By assuming a form for the particular solution, you substitute it back into the equation to determine the unknown coefficients. The goal is to find a specific function that satisfies the non-homogeneous part of the equation.
In the problem, the non-homogeneous part is \( 8x^6 \), prompting us to assume a particular solution of the form \( y_p = Ax^6 \). Upon substitution and simplification, you solve for \( A \) such that the equation balances, yielding a particular solution. This method is usually effective when the non-homogeneous part is a polynomial, exponential, sine, or cosine function.