Problem 30 - Advanced Engineering Mathematics

Question

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

The solution consists of homogeneous and particular parts, solved with initial conditions: plot after determining constants.

Step 1: Identify the Type of Differential Equation

Given the differential equation \( x^2 y'' - 5x y' + 8y = 8x^6 \), we can see it is a non-homogeneous second-order linear differential equation with variable coefficients.

Step 2: Change of Variable

To simplify, use the substitution \( x = e^t \), which implies \( dx/dt = x \). Then, the second derivative with respect to \( x \) becomes the second derivative with respect to \( t \). The transformations are \( y' = x \frac{dy}{dx} = \frac{dy}{dt} \) and \( y'' = \frac{d}{dx}(x \frac{dy}{dx}) = x \frac{d^2y}{dt^2} + \frac{dy}{dt} \). Substitute these into the equation.

Step 3: Convert and Solve the Transformed Homogeneous Part

After substituting the transformations, separate them into the homogeneous equation and solve for complementary function \( y_c \). Ignoring the non-homogeneous term for now, solve the equation \( x \frac{d^2y}{dt^2} - 4 \frac{dy}{dt} + 8y = 0 \). This is a linear differential equation with constant coefficients, solved by assuming \( y = e^{rt} \). Substitute this to find the characteristic equation \( r(r-4)+8=0 \). Find the roots.

Step 4: Solve for the Particular Solution

Solve the non-homogeneous part using undetermined coefficients. Assume a particular solution of a similar form to the non-homogeneous part: \( y_p = Ax^6 \). Differentiate as necessary and substitute into the original non-homogeneous equation to solve for \( A \).

Step 5: Apply the Initial Conditions

Combine the complementary function \( y_c \) and particular solution \( y_p \) to form the general solution. Then, use the initial conditions \( y(\frac{1}{2}) = 0 \) and \( y'(\frac{1}{2}) = 0 \) to solve for the constants within \( y_c \).

Step 6: Graph the Solution

After finding the complete solution, use a graphing utility to plot the curve of the function. Make sure the initial conditions are satisfied on the graph, showing the intersection at \( (\frac{1}{2}, 0) \).

Key Concepts

Non-Homogeneous Differential Equation

Differential equations can be broadly categorized into homogeneous and non-homogeneous equations. In our exercise, we are dealing with a non-homogeneous differential equation. This type of equation includes an additional term on the right-hand side that is not equal to zero. For example, in our given equation, \( x^2 y'' - 5x y' + 8y = 8x^6 \), the term \( 8x^6 \) makes it non-homogeneous.

To solve these equations, we typically aim to find two parts of the solution: the complementary function, which solves the related homogeneous equation (obtained by setting the right-hand side to zero), and the particular solution, which accounts for the non-zero right-hand side.

Understanding the distinction between homogeneous and non-homogeneous equations is crucial as it dictates the methods used to find the solution.

Variable Coefficients

The differential equation in our problem features variable coefficients, meaning that the coefficients of the derivatives are not constants but rather functions of the independent variable, in this case, \( x \). For instance, the term \( -5x y' \) has \( -5x \) as a coefficient, which varies with \( x \).

Variable coefficients often complicate the process of finding the solution compared to constant coefficients. They often require techniques such as substitution or differentiation to simplify the equation, which is why, in this exercise, a change of variable was used by substituting \( x = e^t \) to transform the differential equation into one with constant coefficients.

Initial Conditions

Initial conditions allow us to find a specific solution from the general solution of a differential equation. They provide specific values for the function or its derivatives at a particular point. In our problem, the initial conditions are \( y\left(\frac{1}{2}\right)=0 \) and \( y'\left(\frac{1}{2}\right)=0 \).

These conditions help us determine the constants within the general solution derived from the complementary function and particular solution. Applying initial conditions ensures that the solution not only meets the general form but also aligns precisely with given criteria.

This process of applying initial conditions transforms the general solution into a solution tailored to a specific scenario.

Complementary Function

The complementary function, often denoted as \( y_c \), is the solution to the associated homogeneous differential equation. In our case, we focus on solving \( x^2 y'' - 5x y' + 8y = 0 \), ignoring the non-homogeneous part for the time being.

This function is found using methods like assuming a solution of the form \( y = e^{rt} \), which leads us to the characteristic equation. From there, we determine the roots and construct the solution based on these roots, which may result in exponential, oscillating, or mixed functions depending on whether the roots are real, imaginary, or repeated.

The complementary function is vital as it forms the foundation of the general solution.

Particular Solution

The particular solution, denoted \( y_p \), specifically addresses the non-homogeneous part of the differential equation. In our example, we need to account for the term \( 8x^6 \) present on the right side of the equation.

To find the particular solution, methods such as undetermined coefficients or variation of parameters are employed. In this exercise, a form similar to the non-homogeneous part, \( y_p = Ax^6 \), is assumed and adjusted until it fits the equation when substituted.

This solution, when combined with the complementary function, forms the general solution, fully capturing the behavior of the original differential equation.