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Chapter 4: Problem 29

Find linearly independent functions that are annihilated by the given differential operator. $$(D-6)(2 D+3)$$

Short Answer

Expert verified
The linearly independent functions are \(e^{6x}\) and \(e^{-\frac{3}{2}x}\).

Step by step solution

01

Understand the Differential Operator

The given differential operator is \[(D-6)(2D+3)\]This represents a composition of two first-order linear differential operators: \(D-6\) and \(2D+3\). The operator \(D\) denotes differentiation with respect to some variable, typically \(x\).
02

Analyze the Roots of the Differential Operator

The operator can be rewritten as the composition of two linear operators. We first set each factor equal to zero:\[D - 6 = 0 \quad \Rightarrow \quad D = 6\]\[2D + 3 = 0 \quad \Rightarrow \quad D = -\frac{3}{2}\]These are the roots of the characteristic equation of the differential operator, which indicates the exponential functions involved鈥嬧��.
03

Determine the Linearly Independent Solutions

Each root corresponds to an exponential solution of the differential equation. For a root \(r\), the solution is given by:\[e^{rx}\]Thus, the roots \(6\) and \(-\frac{3}{2}\) yield the solutions:\[e^{6x}\] for \(D = 6\) and \[e^{-\frac{3}{2}x}\]for \(D = -\frac{3}{2}\). These solutions are linearly independent since they involve different exponents.
04

Conclude the Annihilated Functions

The functions that are annihilated by the differential operator \((D-6)(2D+3)\) are \(e^{6x}\) and \(e^{-\frac{3}{2}x}\). These are linearly independent solutions, each corresponding to one root found in Step 2.

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Key Concepts

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

Linearly Independent Functions
Linearly independent functions are an important concept, especially in the context of solving differential equations.
  • To understand linear independence, imagine having two functions that cannot be written as a constant times each other.
  • In mathematical terms, if you have two functions, say \( f(x) \) and \( g(x) \), they are linearly independent if there is no nonzero constant \( c \) such that \( f(x) = c \cdot g(x) \).
  • This property is crucial because it ensures that each function contributes uniquely to the solution of a differential equation, leading to a complete and accurate set of solutions.
For our differential operator \[(D-6)(2D+3)\]the solutions \( e^{6x} \) and \( e^{-\frac{3}{2}x} \) are linearly independent. This is due to their distinct exponential components, \( 6x \) and \(-\frac{3}{2}x \). Thus, they cannot be expressed as multiples of each other.
Characteristic Equation
The characteristic equation is a fundamental concept when working with linear differential operators.
  • It arises when you factor a differential operator and set each part to zero. This provides the roots that are crucial to finding solutions.
  • In our case, from the operator \[(D-6)(2D+3)\] we equate each factor to zero, giving us \[D - 6 = 0 \quad \text{and} \quad 2D + 3 = 0\]
  • Solving these gives us the roots \( D = 6 \) and \( D = -\frac{3}{2} \).
These roots are an essential part of finding the solutions to the differential equation because they directly inform the exponential solutions that emerge.
Exponential Solutions
In the context of differential equations, exponential solutions often represent fundamental solutions.
  • Each root from the characteristic equation corresponds to an exponential function of the form \( e^{rx} \), where \( r \) is a root.
  • In our example, the roots \( 6 \) and \( -\frac{3}{2} \) yield the solutions \( e^{6x} \) and \( e^{-\frac{3}{2}x} \), respectively.
  • The power of exponential solutions lies in their ability to neatly encapsulate the dynamics described by differential equations, particularly linear ones.
These solutions are an inherent part of the process of transforming a differential operator into a comprehensible solution form, typically underpinning the behavior of physical systems, such as growth and decay processes.

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Most popular questions from this chapter

Solve the given differential equation by undetermined coefficients.In Problems \(1-26\) solve the given differential equation by undetermined coefficients. $$y^{(4)}+2 y^{\prime \prime}+y=(x-1)^{2}$$

A mathematical model for the position \(x(t)\) of a moving object is \\[\frac{d^{2} x}{d t^{2}}+\sin x=0\\] Use a numerical solver to graphically investigate the solutions of the equation subject to \(x(0)=0, x^{\prime}(0)=x_{1}, x_{1} \geq 0 .\) Discuss the motion of the object for \(t \geq 0\) and for various choices of \(x_{1}\) Investigate the equation \\[\frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}+\sin x=0\\] in the same manner. Give a possible physical interpretation of the \(d x / d t\) term.

Use a CAS as an aid in solving the auxiliary equation. Form the general solution of the differential equation. Then use a CAS as an aid in solving the system of equations for the coefficients \(c_{i}, i=1,2,3,4\) that results when the initial conditions are applied to the general solution. $$\begin{array}{l} 2 y^{(4)}+3 y^{\prime \prime \prime}-16 y^{\prime \prime}+15 y^{\prime}-4 y=0 \\\ y(0)=-2, y^{\prime}(0)=6, y^{\prime \prime}(0)=3, y^{\prime \prime \prime}(0)=\frac{1}{2} \end{array}$$

Solve the given differential equation by undetermined coefficients.In Problems \(1-26\) solve the given differential equation by undetermined coefficients. $$y^{(4)}-y^{\prime \prime}=4 x+2 x e^{-x}$$

Solve the given initial-value problem. $$\begin{aligned} &\frac{d x}{d t}=y-1\\\ &\frac{d y}{d t}=-3 x+2 y\\\ &x(0)=0, y(0)=0 \end{aligned}$$
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