Problem 12 - Differential Equations and Linear Algebra

Question

Determine the annihilator of the given function. $$F(x)=x \cos 3 x$$.

Step-by-Step Solution

Answer

The annihilator of the given function is \(L = D \cdot D^2\), where \(D\) is the differential operator.

Step 1: Identify the distinct functions

We can identify the distinct functions in the given function: \(F(x) = x \cos{3x}\). Here, we have two distinct functions: \(f_1(x) = x\) and \(f_2(x) = \cos{3x}\).

Step 2: Find annihilators for each distinct function

First, we will find annihilators for each of the distinct functions. For \(f_1(x) = x\), we can find its annihilator by using the differential operator \(D\), where \(D\) is the operator for taking the derivative of a function. The annihilator for \(f_1(x)\) is: \(D \cdot x = \frac{d}{dx}(x) = 1\) For \(f_2(x) = \cos{3x}\), since it's a trigonometric function, we can find its annihilator by using the differential operator \(D^2\), where \(D^2\) is the operator for taking the 2nd derivative of a function. The annihilator for \(f_2(x)\) is: \(D^2 \cdot \cos{3x} = \frac{d^2}{dx^2}(\cos{3x}) = -9 \cos{3x}\)

Step 3: Find the annihilator of the given function

Since we are given a function that is the product of two distinct functions, we can find the annihilator for the given function by taking the product of the annihilators for the two distinct functions. The annihilator for the function \(F(x) = x \cos{3x}\) is: \(L = D \cdot D^2\) To find the differential operator \(L\), we first apply the operator \(D\) to the function \(F(x)\). We have: \(D [F(x)] = D [x \cos{3x}] = x (-3\sin{3x}) + \cos{3x}\) Next, we apply the operator \(D^2\) on the result obtained from the previous step: \(D^2 [D [F(x)]] = D^2 [x (-3\sin{3x}) + \cos{3x}] \) \(= -6x\cos{3x} - 9x\sin{3x} - 3\sin{3x}\) Now, we can apply the annihilator \(L = D \cdot D^2\) to the given function : \(L[F(x)] = D \cdot D^2 [x \cdot \cos{3x}] = -6x\cos{3x} - 9x\sin{3x} - 3\sin{3x}\) So, the annihilator of the given function \(F(x)=x \cos{3x}\) is \(L = D \cdot D^2\).

Key Concepts

Annihilator Method

The Annihilator Method is a systematic approach used in differential equations to find a function that reduces to zero when combined with a certain differential operator. This method can be especially useful for finding particular solutions to non-homogeneous differential equations. It works by identifying functions that, when operated upon by the differential operator, result in the zero function.

• First, distinct functions within a given function are identified.
• Next, annihilators for each of these functions are found.
• The annihilator for the combined function is then determined by integrating these individual annihilators.

Understanding and applying the Annihilator Method can simplify solving differential equations, particularly when dealing with complex functions.

Differential Operators

Differential operators are fundamental tools in calculus that involve differentiation. Essentially, they describe operations like taking derivatives of functions. These operators are symbolized, for example, by \(D\), signifying the derivative with respect to a variable, typically \(x\). Higher-order derivatives use notations like \(D^2\) for second derivatives, \(D^3\) for third derivatives, and so on.

For instance, applying \(D\) to a function \(f(x)\) gives \(Df(x) = \frac{df}{dx}\). Meanwhile, \(D^2\) results in the second derivative: \(D^2f(x) = \frac{d^2f}{dx^2}\). These operators play a critical role in constructing and solving differential equations, making them indispensable in fields such as physics and engineering.

Trigonometric Functions

Trigonometric functions, like \(\cos\), \(\sin\), and \(\tan\), are functions related to angles and periodic phenomena. They are massively important in mathematics due to their periodic properties and their application across various domains, including engineering, physics, and even economics.

In the context of differential equations, trigonometric functions often appear in scenarios involving oscillatory behavior, such as waves or vibrating systems. Thus, understanding how to differentiate and integrate these functions is crucial. For example, the derivative of \(\cos(3x)\) involves using the chain rule, resulting in \(-3\sin(3x)\), demonstrating how differentiation can affect the behavior of these functions.

Product of Functions

When dealing with the product of functions in calculus, one often uses the product rule for differentiation. This is because the derivative of a product is not simply the product of the derivatives. The product rule states that for two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is:\[ (uv)' = u'v + uv'\] Applying this rule is essential when working on problems involving a product of two functions.

For example, in the case of \(x \cos{3x}\), the product rule reveals different derivatives leading to a nuanced function behavior. By accurately applying this rule, solutions to differential equations involving products can be derived efficiently without errors. Understanding these rules enables clear resolution of complex mathematical problems.