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Chapter 4: Problem 10

Given that \(y=c_{1} x^{2}+c_{2} x^{4}+3\) is a two-parameter family of solutions of \(x^{2} y^{\prime \prime}-5 x y^{\prime}+8 y=24\) on the interval \((-\infty, \infty)\), determine whether a member of the family can be found that satisfies the boundary conditions (a) \(y(-1)=0, y(1)=4\) (b) \(y(0)=1, y(1)=2\) (c) \(y(0)=3, y(1)=0\) (d) \(y(1)=3, y(2)=15\).

Short Answer

Expert verified
(a) No, (b) No, (c) Yes, (d) Yes.

Step by step solution

01

Set up the Family Function

The given family of solutions is \(y = c_{1} x^{2} + c_{2} x^{4} + 3\). We need to verify if there exist constants \(c_1\) and \(c_2\) for which the family satisfies different boundary conditions.
02

Analyze Boundary Condition (a)

Substitute \(x = -1\) and \(x = 1\) into the family function for \(y(-1) = 0\) and \(y(1) = 4\), obtaining two equations: 1. \(c_1 + c_2 + 3 = 0\) 2. \(c_1 + c_2 + 3 = 4\).These equations are inconsistent, meaning no \(c_1\) and \(c_2\) satisfy the conditions simultaneously.
03

Analyze Boundary Condition (b)

Substitute \(x = 0\) and \(x = 1\) into the family function for \(y(0) = 1\) and \(y(1) = 2\), resulting in:1. \(3 = 1\), which is inconsistent by itself.Thus, there are no such \(c_1\) and \(c_2\) that satisfy these conditions.
04

Analyze Boundary Condition (c)

Substitute \(x = 0\) and \(x = 1\) into the family function for \(y(0) = 3\) and \(y(1) = 0\):1. \(3 = 3\), identity holds.2. \(c_1 + c_2 + 3 = 0\), which implies \(c_1 + c_2 = -3\).A solution here is \(c_1 = -3 - c_2\), meaning there are infinite solutions.
05

Analyze Boundary Condition (d)

Substitute \(x = 1\) and \(x = 2\) into the family function for \(y(1) = 3\) and \(y(2) = 15\):1. \(c_1 + c_2 + 3 = 3\), simplify to \(c_1 + c_2 = 0\).2. \(4c_1 + 16c_2 + 3 = 15\), simplify to \(4c_1 + 16c_2 = 12\).Solving the equations from the two conditions gives \(c_1 = -3, c_2 = 3\) as a potential solution.

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Key Concepts

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

Second-order differential equations
Second-order differential equations are fundamental in understanding many physical phenomena. These types of equations generally involve the highest derivative being second-order. In mathematical terms, they are written as:
  • For example, in the form: \( ax^2 y'' + by' + cy = g(x) \), where \( y'' \) is the second derivative of \( y \).
  • These equations are important because they can describe processes involving acceleration, forces, and vibrations, among others.

To solve a second-order differential equation, suitable methods include finding a general solution for the homogeneous equation and then a particular solution. Boundary conditions help determine specific solutions from a general solution.
Generally, these equations might have applications in engineering, physics, and other fields related to technology and the natural sciences.
Two-parameter family of solutions
A two-parameter family of solutions involves finding solutions to differential equations involving two arbitrary constants. Here, the constants are often denoted as \( c_1 \) and \( c_2 \). They provide the flexibility needed to meet specific initial or boundary conditions imposed on the problem.
  • In the given exercise, the family of solutions is expressed as \( y = c_1 x^2 + c_2 x^4 + 3 \).
  • The goal is often to determine these parameters so that the given boundary conditions are satisfied.

Such families allow for a wide range of potential solutions, making them versatile.Ultimately, by adjusting \( c_1 \) and \( c_2 \), it is possible to find different curves that the family equation describes.
Boundary conditions
Boundary conditions are specific criteria used to identify unique solutions from a family of potential solutions to a differential equation. They typically set the value of a function or its derivatives at specific points.
Boundary conditions are crucial because:
  • They help transform a general solution into a particular solution, making it applicable to a real-world problem.
  • In the exercise, various boundary conditions are tested to check the viability of solutions.

For instance, given conditions like \( y(-1) = 0 \) and \( y(1) = 4 \), one substitutes these into the family function to derive equations that \( c_1 \) and \( c_2 \) need to satisfy.However, as seen in the solution, not all boundary conditions are possible (as some may lead to inconsistencies).
Inconsistent equations
Inconsistent equations occur when no values satisfy the given system of equations simultaneously. This scenario often becomes evident when testing boundary conditions yields contradictions. Here鈥檚 why it matters:
  • Inconsistent equations indicate that a given set of boundary conditions doesn鈥檛 work with the proposed solution family.
  • For example, the pair of equations \( c_1 + c_2 + 3 = 0 \) and \( c_1 + c_2 + 3 = 4 \) contradict each other. Such contradictions signal inconsistency.

This highlights the importance of checking for consistency, especially when determining specific solutions from a set. If inconsistencies arise, different parameters or initial conditions might need to be considered.

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Most popular questions from this chapter

Find an interval around \(x=0\) for which the given initial-value problem has a unique solution. $$ y^{\prime \prime}+(\tan x) y=e^{t}, \quad y(0)=1, y^{\prime}(0)=0 $$

Determine whether the given functions are linearly independent or dependent on \((-\infty, \infty)\). $$ f_{1}(x)=\cos 2 x, \quad f_{2}(x)=1, \quad f_{2}(x)=\cos ^{2} x $$

Solve the given differential equation by undetermined coefficients. $$ y^{\prime \prime}-y^{\prime}=-3 $$

Solve the given differential equation. \(x^{2} y^{\prime \prime}-x y^{\prime}+2 y=0\)

Solve the given differential equation by undetermined coefficients. $$ y^{\prime \prime}-5 y^{\prime}=2 x^{3}-4 x^{2}-x+6 $$
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