Problem 5 - Calculus: Early Transcendentals

Question

Which of the following functions are solutions of the differential equation \( y^{"} + y = \sin x ? \) (a) \( y = \sin x \) (b) \( y = \cos x \) (c) \( y = \frac {1}{2} x \sin x \) (d) \( y = - \frac{1}{2} x \cos x \)

Step-by-Step Solution

Answer

Function (c) \( y = \frac{1}{2}x\sin x \) is a solution.

Step 1: Verify function (a)

Take the second derivative of the function \( y = \sin x \). The first derivative is \( y' = \cos x \), and the second derivative is \( y'' = -\sin x \). Apply the equation: \( y'' + y = -\sin x + \sin x = 0 eq \sin x \). Thus, \( y = \sin x \) is not a solution.

Step 2: Verify function (b)

Take the second derivative of the function \( y = \cos x \). The first derivative is \( y' = -\sin x \), and the second derivative is \( y'' = -\cos x \). Apply the equation: \( y'' + y = -\cos x + \cos x = 0 eq \sin x \). Thus, \( y = \cos x \) is not a solution.

Step 3: Verify function (c)

Take the second derivative of the function \( y = \frac{1}{2}x\sin x \). First derivative: \( y' = \frac{1}{2}\sin x + \frac{1}{2}x\cos x \). Second derivative: \( y'' = \cos x - \frac{1}{2}x\sin x + \frac{1}{2}\cos x \). Apply the equation: \( y'' + y = (\cos x + \frac{1}{2}\cos x - \frac{1}{2}x\sin x) + \frac{1}{2}x\sin x = \sin x \). Thus, \( y = \frac{1}{2}x\sin x \) is a solution.

Step 4: Verify function (d)

Take the second derivative of the function \( y = -\frac{1}{2}x\cos x \). First derivative: \( y' = \frac{1}{2}\cos x + \frac{1}{2}x\sin x \). Second derivative: \( y'' = -\sin x + \frac{1}{2}x\cos x + \frac{1}{2}\sin x \). Apply the equation: \( y'' + y = (-\sin x + \frac{1}{2}x\cos x + \frac{1}{2}\sin x) + (-\frac{1}{2}x\cos x) = -\sin x \). Thus, \( y = -\frac{1}{2}x\cos x \) is not a solution.

Key Concepts

Second Derivative

Understanding the second derivative is crucial when working with differential equations. The second derivative, denoted as \( y'' \), is the derivative of the first derivative \( y' \). It reveals the rate at which the first derivative itself changes. This can be important in describing the behavior of functions

• The first derivative \( y' \) tells you the slope or rate of change of the original function \( y \).
• The second derivative \( y'' \) gives insights into the curvature or concavity of the function.
• If \( y'' > 0 \), the graph is concave up; if \( y'' < 0 \), it is concave down.

In our exercise, finding the second derivative allows us to test each function against the differential equation \( y'' + y = \sin x \). This informs us if a function behaves in a way that satisfies or fits the conditions of the differential equation. Analyzing the second derivative helps us identify when the solution has a specific curvature that meets the necessary criteria.

Function Verification

Verifying a function as a solution to a differential equation requires substituting the function into the equation to see if it holds true. In the given exercise, a function \( y \) is a solution to the differential equation \( y'' + y = \sin x \) if it satisfies the equation after substituting its derivatives.

• Take the given function and compute its first derivative \( y' \), then the second derivative \( y'' \).
• Substitute \( y'' \) and \( y \) back into the differential equation.
• If both sides of the equation equal, the function is verified as a solution.

For example, function (c) \( y = \frac{1}{2}x\sin x \) was shown to be a solution because its derivatives, when substituted, simplify to \( \sin x \). It demonstrates how crucial correct algebraic manipulation is in the verification process.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) play a significant role in studying differential equations. They exhibit periodic behavior essential for modeling various physical phenomena.

• \( \sin x \) and \( \cos x \) are fundamental sine and cosine functions, formed as the outputs of trigonometric calculations on angles.
• The derivatives of these functions are interconnected, following patterns: the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \(-\sin x \).
• These functions frequently appear in differential equations due to their properties and ability to model oscillations.

In practice, trigonometric identities and derivatives are invaluable for simplifying components within differential equations. The exercise we explored demonstrates their utility in determining whether specific functions fulfill given equations, like \( y'' + y = \sin x \). This interplay between trigonometric functions and calculus can solve complex, real-world problems related to waves and vibrations.