/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Find a fundamental set of soluti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 32

Find a fundamental set of solutions. $$ (D-2)^{3}(D+1)^{2} D y=0 $$

Short Answer

Expert verified
Answer: The fundamental set of solutions for the given linear differential equation consists of the following functions: $y_1(x) = e^{2x}$, $y_2(x) = xe^{2x}$, $y_3(x) = x^2 e^{2x}$, $y_4(x) = e^{-x}$, and $y_5(x) = x e^{-x}$.

Step by step solution

01

Analyze the given equation

The given linear differential equation is: $$ (D-2)^{3}(D+1)^{2} D y=0 $$ The equation contains repeated roots, specifically the roots 2 and -1, each with multiple orders (3 for 2 and 2 for -1).
02

Write the general form of the solution for repeated roots

For the given equation, the general form of the solution is: $$ y(x) = C_1 e^{2x} + C_2 xe^{2x} + C_3 x^2 e^{2x} + C_4 e^{-x} + C_5 x e^{-x} $$ Where \(C_1\), \(C_2\), \(C_3\), \(C_4\), and \(C_5\) are constants.
03

Identify a fundamental set of solutions

From the general form of the solution, the fundamental set of solutions consists of: 1. \(y_1(x) = e^{2x}\) 2. \(y_2(x) = xe^{2x}\) 3. \(y_3(x) = x^2 e^{2x}\) 4. \(y_4(x) = e^{-x}\) 5. \(y_5(x) = x e^{-x}\) These five functions form a fundamental set of solutions to the given linear differential equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Linear Differential Equation
A linear differential equation is one in which the dependent variable and all its derivatives appear to the first power and are not multiplied together. This type of equation is essential because it allows for methods that can systematically find solutions.
Generally, a linear differential equation can be written in the form:
  • \( L[y] = 0 \)
where \( L \) is a linear operator, involving differential operators like \( D = \frac{d}{dx} \). Understanding the structure of these equations is vital to finding their solutions.
Fundamental Set of Solutions
A fundamental set of solutions refers to a group of solutions to a differential equation that can be used to express any solution to the equation. For a linear differential equation with constant coefficients, the general solution can often be written as a linear combination of a set of independent solutions.
The independence of these solutions is crucial because they form a basis for the solution space. To verify if a set is fundamental, you check if the Wronskian determinant of these solutions is non-zero.
In the given exercise, the functions:
  • \( y_1(x) = e^{2x} \)
  • \( y_2(x) = xe^{2x} \)
  • \( y_3(x) = x^2 e^{2x} \)
  • \( y_4(x) = e^{-x} \)
  • \( y_5(x) = x e^{-x} \)
form the fundamental set of solutions.
Repeated Roots
Repeated roots occur when solving the characteristic equation of a differential equation results in some roots appearing more than once. For example, in the given problem, the roots 2 and -1 each have multiplicities of 3 and 2, respectively.
This happens often with higher-order differential equations and affects the form of the solution.
For a root \( r \) repeated \( n \) times, each solution involves increasing powers of \( x \):
  • \( e^{rx} \)
  • \( xe^{rx} \)
  • ... until
  • \( x^{n-1} e^{rx} \)
This pattern ensures all solutions remain linearly independent.
Solution Form
The solution form of a differential equation can vary depending on the nature of its roots. With repeated roots, as seen in the exercise, the solution adopts a form that accommodates these repetitions by including powers of \( x \) multiplied by the exponential term related to the root.
This leads to solutions such as:
  • \( C_1 e^{2x} + C_2 xe^{2x} + C_3 x^2e^{2x} \) for the root 2
  • \( C_4 e^{-x} + C_5 x e^{-x} \) for the root -1
By understanding the basic form, you can construct a comprehensive solution that includes all possible cases.
Subsequently, any solution to the differential equation is a combination of elements of this form.
Linear Operator Notation
Linear operator notation simplifies the way we express and manipulate differential equations. Here, the operator \( D \) denotes differentiation with respect to \( x \), such that \( Dy = \frac{dy}{dx} \).
In linear differential equations, operations are often compactly represented using these operators, like in our problem:
  • \( (D-2)^3(D+1)^2 D y = 0 \)
This notation allows us to focus on the roots and their multiplicities instead of cumbersome manipulation of derivatives.
It is especially helpful in breaking down and solving complex, higher-order differential equations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ \begin{array}{|l} 4 y^{(4)}+8 y^{\prime \prime \prime}+19 y^{\prime \prime}+32 y^{\prime}+12 y=0, \quad y(0)=3, \quad y^{\prime}(0)=-3, \quad y^{\prime \prime}(0)=-\frac{7}{2}, \\ y^{\prime \prime \prime}(0)=\frac{31}{4} \end{array} $$

Use the method suggested by Exercise 34 to find a particular solution in the form \(y_{p}=\int_{x_{0}}^{x} G(x, t) F(t) d t,\) given the indicated fundamental set of solutions. Assume that \(x\) and \(x_{0}\) are in an interval on which the equation is normal. $$ x^{4} y^{(4)}+6 x^{3} y^{\prime \prime \prime}+2 x^{2} y^{\prime \prime}-4 x y^{\prime}+4 y=F(x) ; \quad\left\\{x, x^{2}, 1 / x, 1 / x^{2}\right\\} $$

An equation of the form $$ a_{0} x^{n} y^{(n)}+a_{1} x^{n-1} y^{(n-1)}+\cdots+a_{n-1} x y^{\prime}+a_{n} y=0, \quad x>0 $$ where \(a_{0}, a_{1}, \ldots, a_{n}\) are constants, is an Euler or equidimensional equation. Show that if $$ x=e^{t} \quad \text { and } \quad Y(t)=y(x(t)) $$ then $$ \begin{aligned} x \frac{d y}{d x} &=\frac{d Y}{d t} \\ x^{2} \frac{d^{2} y}{d x^{2}} &=\frac{d^{2} Y}{d t^{2}}-\frac{d Y}{d t} \\ x^{3} \frac{d^{3} y}{d x^{3}} &=\frac{d^{3} Y}{d t^{3}}-3 \frac{d^{2} Y}{d t^{2}}+2 \frac{d Y}{d t} \end{aligned} $$ In general, it can be shown that if \(r\) is any integer \(\geq 2\) then $$ x^{r} \frac{d^{r} y}{d x^{r}}=\frac{d^{r} Y}{d t^{r}}+A_{1 r} \frac{d^{r-1} Y}{d t^{r-1}}+\cdots+A_{r-1, r} \frac{d Y}{d t} $$ where \(A_{1 r}, \ldots, A_{r-1, r}\) are integers. Use these results to show that the substitution (B) transforms (A) into a constant coefficient equation for \(Y\) as a function of \(t\).

Use the method suggested by Exercise 34 to find a particular solution in the form \(y_{p}=\int_{x_{0}}^{x} G(x, t) F(t) d t,\) given the indicated fundamental set of solutions. Assume that \(x\) and \(x_{0}\) are in an interval on which the equation is normal. $$ x y^{(4)}+4 y^{\prime \prime \prime}=F(x) ; \quad\left\\{1, x, x^{2}, 1 / x\right\\} $$

Solve the initial value problem and graph the solution. $$ 4 y^{\prime \prime \prime}-3 y^{\prime}-y=e^{-x / 2}(2-3 x), \quad y(0)=-1, \quad y^{\prime}(0)=15, \quad y^{\prime \prime}(0)=-17 $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.