Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the indicated function is a solution of the given differential equation, that is, substitute the indicated function for y to see that it produces an equality. $$\frac{d^{2} y}{d x^{2}}+y=0 ; y=C_{1} \sin x+C_{2} \cos x$$

Step-by-Step Solution

Answer

The function satisfies the differential equation as substituting it results in a true equality.

Step 1: Differentiate the Function

Given function is \( y = C_1 \sin x + C_2 \cos x \). First, find the first derivative: \( \frac{dy}{dx} = C_1 \cos x - C_2 \sin x \).

Step 2: Find the Second Derivative

Further differentiate the first derivative to obtain the second derivative: \( \frac{d^2 y}{dx^2} = -C_1 \sin x - C_2 \cos x \).

Step 3: Substitute in the Differential Equation

Substitute \( y = C_1 \sin x + C_2 \cos x \) and \( \frac{d^2 y}{dx^2} = -C_1 \sin x - C_2 \cos x \) into the differential equation \( \frac{d^2 y}{dx^2} + y = 0 \).

Step 4: Verify the Equality

Plug the expressions into the equation: \( (-C_1 \sin x - C_2 \cos x) + (C_1 \sin x + C_2 \cos x) = 0 \). Simplify to show: \( 0 = 0 \), which is a true statement.

Key Concepts

Second Derivative

When working with differential equations, you often need to find not just the first derivative, but also the second derivative of a function. The second derivative, denoted as \( \frac{d^2 y}{dx^2} \), gives you insight into the curvature of the function or how it's bending. Think of it as taking the derivative of a derivative.In the given function \( y = C_1 \sin x + C_2 \cos x \), the first step is to differentiate it once to find its first derivative: \( \frac{dy}{dx} = C_1 \cos x - C_2 \sin x \). After that, differentiate this result again to get the second derivative: \( \frac{d^2 y}{dx^2} = -C_1 \sin x - C_2 \cos x \).Why is this process important? The differential equation given, \( \frac{d^2 y}{dx^2} + y = 0 \), specifically requires you to know the second derivative to verify if the original function fits the equation. The second derivative often appears in physics to model systems involving acceleration, such as motion under constant force.

Trigonometric Functions

Trigonometric functions like sine \( \sin \) and cosine \( \cos \) play a pivotal role in solving differential equations, especially those modeling periodic or oscillatory phenomena.

• Basic Definitions: The sine function represents the vertical component of a point on the unit circle, while the cosine function represents the horizontal component.
• Derivative Relations: Derivatives of trigonometric functions have predictable patterns. For example, the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \(-\sin x \).
• Combining Functions: The function \( y = C_1 \sin x + C_2 \cos x \) combines both sine and cosine, making use of their orthogonal properties where \( \sin \) and \( \cos \) have specific symmetry and phase relationships.

Understanding these functions and their properties allows one to manipulate and solve different equations that mimic real-world periodic behaviors, like sound waves or the swinging of a pendulum.

Solution Verification

Verification in differential equations is crucial to confirm whether a proposed solution actually satisfies the equation.To verify, substitute the function and its derivatives back into the differential equation. In this case, we substitute the function \( y = C_1 \sin x + C_2 \cos x \) and its second derivative \( \frac{d^2 y}{dx^2} = -C_1 \sin x - C_2 \cos x \) into the equation \( \frac{d^2 y}{dx^2} + y = 0 \).

• Substitution: When we plug in the second derivative and the original function into the differential equation, you get \((-C_1 \sin x - C_2 \cos x) + (C_1 \sin x + C_2 \cos x) \).
• Simplification: Notice that each term cancels out to leave \(0\). Simplifying gives \(0 = 0\), confirming that the equality holds true.

This step-by-step substitution and checking ensure that we've accurately solved the differential equation and that our proposed solution is correct, validating both mathematically and conceptually the solution found.