Problem 34 - Advanced Engineering Mathematics

Question

Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$ \begin{aligned} &2 x^{2} y^{\prime \prime}+5 x y^{\prime}+y=x^{2}-x \\ &y=c_{1} x^{-1 / 2}+c_{2} x^{-1}+\frac{1}{15} x^{2}-\frac{1}{6} x,(0, \infty) \end{aligned} $$

Step-by-Step Solution

Answer

The given function satisfies the differential equation on \((0, \infty)\).

Step 1: Substitute the General Solution

To verify that the given family of solutions is correct, first substitute the function \( y = c_1 x^{-1/2} + c_2 x^{-1} + \frac{1}{15} x^2 - \frac{1}{6} x \) into the left-hand side of the differential equation. Compute its derivatives \( y' \) and \( y'' \) first.

Step 2: Find the First Derivative

Differentiate \( y \) with respect to \( x \). Using power rule: \[ y' = -\frac{1}{2} c_1 x^{-3/2} - c_2 x^{-2} + \frac{2}{15} x - \frac{1}{6}. \]

Step 3: Find the Second Derivative

Differentiate \( y' \) with respect to \( x \) again to find \( y'' \): \[ y'' = \frac{3}{4} c_1 x^{-5/2} + 2 c_2 x^{-3} + \frac{2}{15}. \]

Step 4: Substitute Derivatives into the Differential Equation

Substitute \( y \), \( y' \), and \( y'' \) into the given differential equation:\[ 2x^{2}(\frac{3}{4}c_1 x^{-5/2} + 2c_2 x^{-3} + \frac{2}{15}) + 5x(-\frac{1}{2}c_1 x^{-3/2} - c_2 x^{-2} + \frac{2}{15}x - \frac{1}{6}) + (c_1 x^{-1/2} + c_2 x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x). \]

Step 5: Simplify and Verify

Simplify the expression from step 4. Each term cancels or reduces appropriately such that the left-hand side simplifies to \( x^2 - x \). This matches the right-hand side, verifying that the given function is indeed a solution to the differential equation in the interval \((0, \infty)\).

Key Concepts

General Solution

In the world of differential equations, the general solution is a crucial concept that provides a broad understanding of potential answers to an equation. For a nonhomogeneous differential equation like the one given, finding the general solution means determining all possible functions that satisfy the equation.
It encompasses both the specific solutions as a response to the inhomogeneous part of the equation and the homogeneous solution related to the equation's natural response without the external influences.
Think of it as having a mix of a particular solution, which directly solves the equation, and a complementary function that reflects the equation’s inherent characteristics. It can be represented as a "two-parameter family of functions." In this example, the form \(y = c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x\) captures every solution possible, where \(c_1\) and \(c_2\) are arbitrary constants that make the solution adaptable.

Two-Parameter Family

The term "two-parameter family" refers to the set of solutions derived from the constants, typically shown as \(c_1\) and \(c_2\), in the general solution of a differential equation.
These constants represent an infinite array of specific solutions that can satisfy the differential equation. By adjusting \(c_1\) and \(c_2\), you can produce numerous valid solutions, each fitting a particular set of initial conditions or additional constraints.
This variability is pivotal, as it provides the flexibility needed for the general solution to fit different scenarios dictated by the problem’s context.

• These parameters give the general solution its "family" designation, as they spawn multiple solutions.
• Each unique combination of \(c_1\) and \(c_2\) results in a distinct function.

Thus, the solution is not a single function but a collection that represents all potential behaviors consistent with the given differential equation.

Verification Process

Verification is about ensuring that the proposed solution satisfies the differential equation from which it was derived. It's a way to confirm that your hard work has paid off and that the solution stands strong against the conditions of the problem.
Here's how the verification process works for our differential equation:

• First, compute the necessary derivatives of the proposed solution, \(y', y''\), to plug them back into the original equation.
• Substitute each of these derivatives and the function itself into the equation. For our exercise, this means placing each derivative and the function into the left-hand side of the equation \(2x^{2}y'' + 5xy' + y\).
• Simplify the expression to check whether it equals the right side, \(x^2 - x\).

If it balances out correctly, as in this case, it means you have successfully verified that your function satisfies the differential equation across the specified interval \((0, \infty)\). Full verification is often a satisfying confirmation of a solution!

Second-Order Differential Equation

Differential equations can vary in complexity, and a second-order differential equation typically involves the second derivative of the dependent variable with respect to the independent variable, which is exemplified by the term \(y''\) in our case.
These equations are vital in modeling a multitude of physical, social, and economic processes.

• The equation involves solving for a function of which both the value and the slope (first derivative) can affect the acceleration or the rate of change of the slope (second derivative).
• Second-order differential equations often appear in physics in contexts such as motion, where acceleration is directly related to force.

In solving them, different methods apply, ranging from reduction of order, variation of parameters, or utilizing an auxiliary equation approach. For our exercise, the equation \(2x^{2} y'' + 5x y' + y = x^{2} - x\) was addressed by identifying solutions within a form that accommodates the nonhomogeneous aspect
(thanks to the presence of the term \(x^2 - x\)) while also incorporating constants that connect back to the homogeneous part. This combination allows for exploring a full spectrum of solutions.