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Chapter 4: Problem 36

Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$ \begin{aligned} &2 x^{2} y^{\prime \prime}+5 x y^{\prime}+y=x^{2}-x \\ &y=c_{1} x^{-12}+c_{2} x^{-1}+\frac{1}{15} x^{2}-\frac{1}{6} x, \quad(0, \infty) \end{aligned} $$

Short Answer

Expert verified
The proposed function solves the differential equation on \((0, \infty)\).

Step by step solution

01

Identify the Differential Equation and Solution

The differential equation given is \(2x^2y'' + 5xy' + y = x^2 - x\), and the proposed solution is \(y = c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\). We need to check whether this family of functions solves the given differential equation on the interval \((0, \infty)\).
02

Differentiate the Proposed Solution

Find the first and second derivatives of the proposed solution.\[y' = -12c_1x^{-13} - c_2x^{-2} + \frac{2}{15}x - \frac{1}{6}\] \[y'' = 156c_1x^{-14} + 2c_2x^{-3} + \frac{2}{15}\].
03

Substitute Derivatives into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) back into the differential equation:\[2x^2(156c_1x^{-14} + 2c_2x^{-3} + \frac{2}{15}) + 5x(-12c_1x^{-13} - c_2x^{-2} + \frac{2}{15}x - \frac{1}{6}) + c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x = x^2 - x\].
04

Simplify the Substituted Equation

Carry out the multiplication and addition:\[ \begin{aligned}&312c_1x^{-12} + 4c_2x^{-1} + \frac{4}{15}x^2 - 60c_1x^{-12} - 5c_2x^{-1} + \frac{10}{15}x^2 - \frac{5}{6}x \&+ c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x = x^2 - x\end{aligned} \]Combine like terms to simplify.
05

Verify Solution Equality

Verify that the simplified expression matches the nonhomogeneous term \(x^2 - x\). The terms under \(x^{-12}\), \(x^{-1}\), and coefficients of \(x^2\) and \(x\) either cancel each other or match the right side nonhomogeneous term, resulting in\[x^2 - x = x^2 - x\].
06

Conclusion: Validity of the Solution

Since the simplified left-hand side equals the right-hand side \(x^2 - x\), the proposed function \(y = c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\) is indeed a general solution of the differential equation on \((0, \infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonhomogeneous Differential Equations
A nonhomogeneous differential equation is a type of differential equation that contains a non-zero term, making it 'nonhomogeneous.' These equations take the form:
  • The left side consists of derivatives of the function.
  • The right side is a non-zero function, called the nonhomogeneous term.
In the given exercise, the differential equation is \(2x^2y'' + 5xy' + y = x^2 - x\). Here, the nonhomogeneous term is \(x^2 - x\). This means that our equation has some additional component that makes it differ from a purely homogeneous equation. Understanding this component is crucial when finding solutions, as it usually dictates a particular piece of the entire solution.
General Solution
The general solution of a differential equation is an expression that encompasses all possible solutions given certain initial conditions. It includes both the homogeneous solution and a particular solution.
  • The homogeneous solution is what solves the equation when the nonhomogeneous term is set to zero (e.g., \(2x^2y'' + 5xy' + y = 0\)).
  • A particular solution specifically satisfies the entire nonhomogeneous equation, taking the non-zero term into account.
In our exercise, the proposed general solution is \(y = c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\), where \(c_1\) and \(c_2\) are constants. The first two terms constitute the solution to the homogeneous part, while the remaining fit the nonhomogeneous equation.
Parameter Family of Functions
When dealing with differential equations, the solution often involves a parameter family of functions. This concept means that the solution is not a single function but a whole family of functions parameterized by constants.
The constants, like \(c_1\) and \(c_2\) in our exercise, allow the solution to apply to a range of initial conditions or scenarios. These constants are critical because:
  • They account for the infinite possibilities inherent in solving a differential equation.
  • They are determined by initial or boundary conditions outside the explicit form of the solution.
Thus, in the given solution \(y = c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\), \(c_1\) and \(c_2\) represent the adjustable aspects of the function family that constitute a general solution.
Interval Verification
Verifying a solution over an interval means ensuring that the proposed solution satisfies the differential equation for all values within that specified interval. This step is crucial, as solutions can be valid over some intervals and not others.
In the exercise, the interval given is \((0, \infty)\). This particular interval implies:
  • The function and its derivatives must be defined and differentiable over this range.
  • Computations show that by substituting back into the differential equation, the left side equals the right side consistently across the interval.
Thus, the verification process confirms that for all positive \(x\), the proposed solution \(y = c_1x^{-12} + c_2x^{-1} + \frac{1}{15}x^2 - \frac{1}{6}x\) actually resolves the equation over \((0, \infty)\). This solidifies the validity of the general solution for the specified interval.

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Most popular questions from this chapter

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