Problem 20 - A First Course in Differential Equations with Modeling Applications

Question

Find a linear differential operator that annihilates the given function. $$1+\sin x$$

Step-by-Step Solution

Answer

The operator \((D^2 + 1)\) annihilates the function \(1 + \sin x\).

Step 1: Understand the Function

The given function is \(f(x) = 1 + \sin x\). This function consists of a constant term, 1, and a sinusoidal term, \(\sin x\).

Step 2: Identify Annihilators

The annihilator of a function is a differential operator that, when applied to the function, results in zero. For a constant function, such as 1, the basic differential operator that annihilates it is the zero-order operator, which is essentially no differentiation, leading to \(\frac{d}{dx}(1) = 0\). For \(\sin x\), the differential operator \(D^2 + 1\), where \(D = \frac{d}{dx}\), will serve as an annihilator. This is because \(\frac{d^2}{dx^2}(\sin x) + \sin x = 0\).

Step 3: Combine Annihilators

Since the given function is a linear combination of \(1\) and \(\sin x\), we need a single operator that will annihilate the entire expression. The least common multiple of the operators for each part is used to achieve this. An operator that annihilates both 1 and \(\sin x\) can be constructed by using the operator that annihilates \(\sin x\), which is \((D^2 + 1)\), since it also annihilates the constant term by default.

Step 4: Write the Linear Differential Operator

Thus, the linear differential operator that annihilates the function \(1 + \sin x\) is \((D^2 + 1)\). When this operator is applied to the function \(1 + \sin x\), it results in zero, confirming it as the correct annihilator.

Key Concepts

Linear Differential Operators

In the realm of differential equations, a linear differential operator is an essential concept. Essentially, it is a function that applies differentiation to other functions. These operators are usually denoted using derivatives, such as \( D = \frac{d}{dx} \). A clear understanding of these operators is crucial, as they allow us to transform problems into manageable equations.

• Linear differential operators have the property of linearity, meaning they adhere to linear algebraic rules like addition and scalar multiplication.
• For example, if \( D_1 \) and \( D_2 \) are linear operators and \( a \) is a constant, then the operator \( aD_1 + D_2 \) will also be linear.
• These operators are pivotal in solving differential equations as they help in simplifying complex expressions.

Linear differential operators are not just limited to simple differentiation. They can involve higher-order derivatives, like the second derivative \( D^2 = \frac{d^2}{dx^2} \), forming part of more powerful tools to solve differential equations. Understanding how these operators interact with functions opens the door to efficiently solve or simplify problems in mathematical analysis and engineering applications.

Annihilator Method

The annihilator method is a powerful technique widely used for solving differential equations. This method involves finding a differential operator, known as an annihilator, that turns a given function into zero.

• To solve using the annihilator method, you start by identifying a differential operator that, when applied, makes the function disappear by reducing it to zero.
• The process usually involves combining annihilators of individual components of the function. If the function is a combination of terms, like in \( f(x) = 1 + \sin x \), you need to find individual annihilators first.

For example, the annihilator for \( \sin x \) is \( D^2 + 1 \). Applying this operator indicates that when you take the second derivative and add the original function, it results in zero. Therefore, this operator also, by default, works for a constant like 1 as constants have zero derivatives. Thus, when a function consists of multiple terms, the operator for the most complex term typically annihilates the rest.
In practical terms, this method structure becomes very handy. It simplifies the process of finding the complete solution for complex differential equations by reducing sub-problems into simpler, manageable parts.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) play a significant role in calculus and differential equations. These functions, due to their periodic nature, have specific properties that interact uniquely with differential operators.

• The derivatives of trigonometric functions are cyclical; for instance, repeatedly differentiating \( \sin x \) cycles through \( \cos x \), \( -\sin x \), and \( -\cos x \).
• This cyclical property leads to the creation of differential operators that annihilate these functions.

For any trigonometric function, there's a unique combination of derivatives that can result in the function being annihilated. In our example of \( \sin x \), the operator \( D^2 + 1 \) annihilates the function completely by producing zero.
Understanding these properties is crucial when dealing with trigonometric functions in differential equations, as it allows us to find solutions more easily.
By playing around with derivatives of trigonometric functions, you can discover patterns and apply appropriate operators that simplify the problem-solving process, especially when solving complex trigonometric equations and systems. These insights are frequently used in engineering, physics, and many applied sciences.