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Chapter 4: Problem 16

Find a linear differential operator that annihilates the given function. $$ x^{3}(1-5 x) $$

Short Answer

Expert verified
The operator that annihilates the function is \((D^4)(D-5)\).

Step by step solution

01

Identify the Polynomial Components

The given function is the product of two polynomials: the first is simple, \(x^3\), and the second is \((1 - 5x)\), which is a first - degree polynomial.
02

Annihilate Each Component Separately

To find an annihilator for \(x^3\), note that the differential operator \((D^4)\) will annihilate \(x^3\). This means that applying \(D^4\) to \(x^3\) results in zero.
03

Annihilate the Linear Factor

For the linear component \((1 - 5x)\), we use the operator \((D - 5)\). This operator will annihilate a polynomial of the form \((1 - 5x)\), meaning that applying \((D-5)\) to it will result in zero.
04

Combine Annihilators

To annihilate the entire product \(x^3 (1 - 5x)\), multiply the annihilators from steps 2 and 3. Therefore, the linear differential operator is given by \((D^4)(D-5)\). This combined operator will annihilate the entire polynomial.

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Key Concepts

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

Annihilator
In differential equations, an annihilator is a linear differential operator that, when applied to a function, results in zero. This means it "annihilates" the function. In practice, finding an annihilator involves determining a differential operator that can simplify a given function to zero.

When dealing with polynomial functions, each part can have its own annihilator. For example, the polynomial function x^3(1-5x) involves two sections: x^3 and (1 - 5x) . Each section is annihilated separately, then combined. The goal is to create a singular operator that applies to the entire function, simplifying calculations in solving differential equations.
Polynomial Components
Polynomial components refer to individual elements that make up a polynomial function. In the example x^3(1-5x) , the polynomial components are x^3 and (1 - 5x) .

This function is a product of these two terms, where x^3 is a third-degree term, and (1-5x) is a first-degree polynomial. Understanding each component is crucial since it allows us to tackle each separately when finding an annihilator.
  • x^3: A simple term representing a cubic polynomial.
  • 1-5x: A linear polynomial of first-degree.
Recognizing these components helps in setting up a logical pathway to find the annihilator for the entire function.
Differential Operator
A differential operator is a tool in calculus represented often by D , signifying differentiation concerning a variable, like x . It is used to compute derivatives of functions, which is essential in finding annihilators.

For example, D^4 refers to taking the fourth derivative of a function regarding x . In the polynomial context, D^4 acts on x^3 to produce zero. Thus, it annihilates it.
  • D^n: Represents the n-th differentiation.
  • D-5: An operator used to simplify functions of the form (1-5x) , leading it to zero.
Learning how to apply differential operators effectively helps in solving differential equations by simplifying and reducing complex polynomials to zero.
Step-by-Step Solution
A step-by-step solution breaks down the process of solving a problem into clear, manageable parts. For our exercise, understanding how to find an annihilator involves such detailed steps.

Step 1: Identify Polynomial Components
First, look at the function to find its polynomial parts, x^3 and 1-5x .

Step 2: Annihilate Each Component
Determine how to annihilate x^3 and (1-5x) separately. Use D^4 for x^3 , leading it to zero. For (1-5x) , apply (D-5) .

Step 3: Combine Annihilators
Finally, to annihilate the entire product x^3 (1-5x) , multiply the individual operators, creating (D^4)(D-5) .

By following these steps, you can systematically tackle the problem and ensure comprehensive understanding.

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Most popular questions from this chapter

Solve each differential equation by variation of parameters. State an interval on which the general solution is defined. $$y^{\prime \prime \prime}+4 y^{\prime}=\sec 2 x$$

Write the given differential equation in the form \(L(y)=g(x)\), where \(L\) is a linear differential operator with constant coefficients. If possible, factor \(L\). $$ y^{(4)}+8 y^{\prime}=4 $$

Solve the given differential equation by using the substitution \(u=y^{\prime}\). $$ y^{\prime \prime}=1+\left(y^{\prime}\right)^{2} $$

Write the given differential equation in the form \(L(y)=g(x)\), where \(L\) is a linear differential operator with constant coefficients. If possible, factor \(L\). $$ y^{\prime \prime}-4 y^{\prime}-12 y=x-6 $$

Determine whether the given functions are linearly independent or dependent on \((-\infty, \infty)\). $$ f_{1}(x)=x, \quad f_{2}(x)=x-1, \quad f_{3}(x)=x+3 $$
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