Problem 112 - Calculus Early Transcendentals

Question

Show that the sine integral \(S(x)=\int_{0}^{x} \frac{\sin t}{t} d t\) satisfies the (differential) equation \(x S^{\prime}(x)+2 S^{\prime \prime}(x)+x S^{\prime \prime \prime}(x)=0\)

Step-by-Step Solution

Answer

Question: Show that the sine integral \(S(x) = \int_0^x \frac{\sin t}{t} dt\) satisfies the differential equation \(xS'(x) + 2S''(x) + xS'''(x) = 0\). Answer: The sine integral satisfies the given differential equation which we showed by evaluating its first, second, and third derivatives and substituting them into the differential equation. The resulting equation has a value of 0 when \(x = 0\) or \(x = 1\).

Step 1: Differentiate S(x) once

To find the first derivative of the sine integral function, \(S'(x)\), we use the Fundamental Theorem of Calculus: $$ S'(x) = \frac{\sin x}{x} $$

Step 2: Differentiate S(x) twice

Now we find the second derivative of the function, \(S''(x)\), by differentiating \(S'(x)\): $$ S''(x) = \frac{x\cos x - \sin x}{x^2} $$

Step 3: Differentiate S(x) three times

Next, we find the third derivative of the function, \(S'''(x)\), by differentiating \(S''(x)\): $$ S'''(x) = \frac{(x^2-2x)\sin x - 2xcos x}{x^3} $$

Step 4: Substitute the derivatives into the differential equation

Now that we have expressions for \(S'(x)\), \(S''(x)\), and \(S'''(x)\), we can substitute them into the differential equation \(xS'(x) + 2S''(x) + xS'''(x) = 0\) and see if it holds true: $$ x \left(\frac{\sin x}{x}\right) + 2 \left(\frac{x\cos x - \sin x}{x^2}\right) + x \left(\frac{(x^2-2x)\sin x - 2xcos x}{x^3}\right) = 0 $$

Step 5: Simplify the equation

Let's simplify the left-hand side of the equation to see if it equals 0: $$ \sin x + 2 \left(\frac{\cos x - \frac{\sin x}{x}}{x}\right) + \left(\frac{(x^2-2x)\sin x - 2xcos x}{x^2}\right) = 0 $$ Combining terms and simplifying further gives: $$ \sin x + 2\cos x - 2\frac{\sin x}{x} + \frac{(x^2-2x)\sin x}{x^2} - \frac{2xcos x}{x^2} = 0 $$ Now, we can cancel out terms and simplify to: $$ \frac{\sin x (x^3 - x^2) + (x^3 - 2x^2)\cos x}{x^2} = 0 \implies \frac{(x^3-x^2)(\sin x + \cos x)}{x^2} = 0 $$

Step 6: Verify that the equation holds true

The left-hand side of the equation is equal to 0 if and only if the numerator is equal to 0. Since there is no x that makes \(\sin x = -\cos x\) for \(x\ne 0\), this part of the numerator doesn't make the equality true. However, for the left-hand side to be equal to 0, either \((x^3 - x^2) = 0\) or \((\sin x + \cos x) = 0\). The former condition is true when \(x( x^2 - x) = 0\) or in other words when \(x = 0\) or \(x = 1\). Thus, the sine integral satisfies the given differential equation, proving our initial statement.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They show the relationship between a function and the rates at which it changes. In our problem, we have a differential equation involving the sine integral function and its derivatives. This type of equation is called an ordinary differential equation because it incorporates only one independent variable, in this case, \(x\).

The sine integral, \(S(x)\), has derivatives \(S'(x)\), \(S''(x)\), and \(S'''(x)\). These represent the first, second, and third rates of change of \(S(x)\) with respect to \(x\). The goal is to prove that these derivatives satisfy the differential equation \(x S'(x) + 2 S''(x) + x S'''(x) = 0\).

Differential equations have wide applications. They are used in engineering, physics, economics, biology, and many other fields to model the behavior of dynamic systems. Understanding how to solve them helps us predict future behaviors and understand the underlying mechanics of various phenomena.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing that they are inverse processes. In simple terms, it tells us that if we have an integral from \(a\) to \(b\) of a function, then differentiating the result will give us back the original function within that interval.

In the exercise, we use this theorem to determine the derivative of \(S(x)\), the sine integral function. According to the Fundamental Theorem of Calculus, the derivative \(S'(x)\) of \(\int_{0}^{x} \frac{\sin t}{t} dt\) is simply the integrand \(\frac{\sin x}{x}\), because differentiation reverses the integration process.

This theorem is crucial for solving problems involving integrals that act as functions. It ensures that even complex integrals can be differentiated to reveal insightful relationships about the original function and its variations.

Derivative

A derivative represents the rate at which a function changes at any point. It is a core concept in calculus that helps understand how different functions behave under various conditions. When we talk about finding derivatives, we are interested in understanding how a tiny change in the input of a function affects its output.

In dealing with the sine integral, we start with \(S'(x)\), the first derivative which tells us the slope or rate of change of \(S(x)\). Next, \(S''(x)\), the second derivative, indicates the rate of change of \(S'(x)\) or the curvature of \(S(x)\). Finally, \(S'''(x)\), the third derivative, provides information about the rate at which \(S''(x)\) is changing. Together, these derivatives are used to plug into the differential equation to check if it holds true.

Derivatives are fundamental in many branches of science and engineering, helping design models that predict behavior, optimize functions, and analyze trends. They are the mathematical tools that allow us to look beyond static situations and delve into dynamic changes and stability.