Problem 24 - Advanced Engineering Mathematics

Question

Verify that the indicated family of functions is a solution to the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ \begin{aligned} &x^{3} \frac{d^{3} y}{d x^{3}}+2 x^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=12 x^{2} \\ &y=c_{1} x^{-1}+c_{2} x+c_{3} x \ln x+4 x^{2} \end{aligned} $$

Step-by-Step Solution

Answer

The given family of functions is a solution to the differential equation.

Step 1: Find the Derivatives

We'll begin by calculating the first, second, and third derivatives of the given function.Given \( y = c_1 x^{-1} + c_2 x + c_3 x \ln x + 4x^2 \).First derivative: \[ \frac{dy}{dx} = -c_1 x^{-2} + c_2 + c_3 (\ln x + 1) + 8x \]Second derivative:\[ \frac{d^2 y}{dx^2} = 2c_1 x^{-3} + \frac{c_3}{x} + 8 \]Third derivative:\[ \frac{d^3 y}{dx^3} = -6c_1 x^{-4} - \frac{c_3}{x^2} \].

Step 2: Substitute Derivatives into Differential Equation

Substitute the derivatives found in Step 1 into the given differential equation:\[ x^3 \frac{d^3 y}{dx^3} + 2x^2 \frac{d^2 y}{dx^2} - x \frac{d y}{dx} + y = 12x^2 \].Let's substitute:- \( x^3 \frac{d^3 y}{dx^3} = x^3 (-6c_1 x^{-4} - \frac{c_3}{x^2}) = -6c_1 x^{-1} - c_3 x \)- \( 2x^2 \frac{d^2 y}{dx^2} = 2x^2 (2c_1 x^{-3} + \frac{c_3}{x} + 8) = 4c_1 x^{-1} + 2c_3 x + 16x^2 \)- \( -x \frac{dy}{dx} = -x(-c_1 x^{-2} + c_2 + c_3 (\ln x + 1) + 8x) = c_1 x^{-1} - c_2 x - c_3 x \ln x - c_3 x - 8x^2 \)

Step 3: Simplify the Left-Hand Side

Combine the substituted terms from Step 2.Combine:\[ -6c_1 x^{-1} - c_3 x + 4c_1 x^{-1} + 2c_3 x + 16x^2 + c_1 x^{-1} - c_2 x - c_3 x \ln x - c_3 x - 8x^2 + c_1 x^{-1} + c_2 x + c_3 x \ln x + 4x^2 \]The terms simplify to:- The \( c_1 x^{-1} \) terms combine to zero- The \( c_2 x \) terms combine to zero- The \( c_3 x \ln x \) terms cancel- The \( c_3 x \) terms cancel- The \( 16x^2 - 8x^2 + 4x^2 \) simplifies to \(12x^2\)

Step 4: Verification

After simplification, the non-cancelled terms in the left-hand side add up to \(12x^2\).Thus, the left-hand side matches the right-hand side of the differential equation, \(12x^2\). This verifies that the given family of functions satisfies the differential equation.

Key Concepts

Higher-Order Derivatives

In the provided problem, understanding higher-order derivatives is crucial. These derivatives, specifically the first, second, and third, are used to solve and verify the differential equation given in the step-by-step solution.

Differential equations involve derivatives of an unknown function, and as you move to higher orders, you're essentially examining the function more deeply.

The first derivative \( \frac{dy}{dx} \) represents the rate of change or the slope of \( y \) at any point. The second derivative \( \frac{d^2 y}{dx^2} \) provides information about the curvature or concavity, indicating how the slope itself is changing. Lastly, the third derivative \( \frac{d^3 y}{dx^3} \) gives insight into the variation rate of the curvature, or how the curvature is evolving.

In the exercise, these derivatives help break down the family of functions we are dealing with, thus simplifying the original equation into terms we can easily manipulate for solution verification.

Family of Solutions

A family of solutions refers to a set of functions that satisfy a given differential equation under various conditions. In the problem, we have the family: \( y = c_1 x^{-1} + c_2 x + c_3 x \ln x + 4x^2 \). Each of these terms is influenced by constants \( c_1, c_2, \) and \( c_3 \), which allow for different solutions within the family based on initial or boundary conditions.

This family represents the broader class of answers we could expect from this differential equation, depending on the aforementioned constants.

By varying these constants, we essentially change the behavior and shape of the curve represented by the function, adjusting it to meet specific criteria or conditions posed by the problem.

For educational purposes, understanding how these constants impact the solution helps in grasping the flexibility and wide range of possible solutions to a differential equation.

Interval of Definition

The interval of definition, \( I \), is the range of the independent variable, here \( x \), for which the solution is valid. This is crucial since certain solutions might not make sense outside of a certain piece of the x-axis.

In solving differential equations, the interval of definition ensures that the solution behaves properly and doesn’t encounter any undefined behavior, such as division by zero or taking the logarithm of a negative number.

For the given function, terms like \( x^{-1} \) and \( x \ln x \) indicate that the interval of definition must exclude zero and must be in the domain of the logarithmic function.

Determining the correct interval of definition is vital, as it might affect the applicability and physical relevance of the solution in practical scenarios.

Verification of Solutions

Verification of solutions involves confirming that the proposed function does, in fact, satisfy the differential equation. This step is crucial as it ensures the solution obtained genuinely represents the system described by the equation.

In the problem, this involves substituting the derivatives back into the original equation and checking if both sides are equal.

The substitution process elucidates whether the algebraic manipulations hold true, confirming the solution matches the predetermined conditions set by the differential equation.

When the left-hand side simplifies exactly to the right-hand side, as shown in the simplified process leading to \( 12x^2 \), it indicates a successful verification.

This not only assures the accuracy of the solution but also enhances understanding of the relationship between the function and its derivatives in context to the equation given.