Problem 4

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. $$ \left(\frac{d y}{d x}\right)^{2}+y^{2}=1 ; y=\sin (x+C) \text { and } y=\pm 1 $$

Step-by-Step Solution

Verified

Answer

Both $ y = \sin(x + C) $ and $ y = \pm 1 $ satisfy the differential equation.

1Step 1: Understand the Function

We are given the function $ y = \sin(x + C) $ and $ y = \pm 1 $. We need to substitute this function into the differential equation $ \left(\frac{dy}{dx}\right)^{2} + y^{2} = 1 $.

2Step 2: Calculate the Derivative

Find the derivative of $ y = \sin(x + C) $. We have $ \frac{dy}{dx} = \cos(x + C) $. For $ y = \pm 1 $, the derivative $ \frac{dy}{dx} = 0 $ since it's a constant.

3Step 3: Substitute the Derivative into the Equation

Substitute $ \frac{dy}{dx} = \cos(x + C) $ into the differential equation: $ (\cos(x + C))^{2} + (\sin(x + C))^{2} = 1 $.

4Step 4: Apply Trigonometric Identity

We know the Pythagorean identity $ \cos^{2}(x + C) + \sin^{2}(x + C) = 1 $ is always true. Therefore, $ \left(\frac{dy}{dx}\right)^{2} + y^{2} = 1 $ holds for $ y = \sin(x + C) $.

5Step 5: Verify for $y = \pm 1$ Case

Substitute $ y = \pm 1 $ into the original differential equation. Since $ \frac{dy}{dx} = 0 $, the equation becomes $ 0^{2} + (\pm 1)^{2} = 1 $, which is true.

Key Concepts

Understanding Trigonometric FunctionsExploring the Pythagorean IdentitySolution Verification Process

Understanding Trigonometric Functions

Trigonometric functions, like sine and cosine, are critical in the study of waves and oscillations. For instance, when dealing with the function $ y = \sin(x + C) $, we observe a sinusoidal pattern. This function represents a wave that oscillates between -1 and 1.
Understanding trigonometric functions and their properties is crucial as they offer solutions to numerous mathematical problems, including differential equations.

The sine function is periodic with a period of $2\pi$.
It has specific values like $ \sin(0) = 0 $, $ \sin(\frac{\pi}{2}) = 1 $, $ \sin(\pi) = 0 $, which help in solving equations and verifying solutions.

Using this information, we can explore various solutions derived from sine functions and apply them effectively to validate the solutions of differential equations.

Exploring the Pythagorean Identity

The Pythagorean identity is a fundamental trigonometric identity that simplifies the evaluation of trigonometric functions in equations. It states that $ \cos^{2}(x) + \sin^{2}(x) = 1 $.
This identity is particularly useful because it holds true for all values of $ x $, facilitating the simplification of complex expressions involving trigonometric functions.
For the given problem, when we substitute $ \frac{dy}{dx} = \cos(x + C) $ and use the identity, we get:

$ (\cos(x + C))^2 + (\sin(x + C))^2 = 1 $

Applying this identity verifies that our differential equation $ \left(\frac{dy}{dx}\right)^2 + y^2 = 1 $ holds true for the function $ y = \sin(x + C) $.
The simplicity and comprehensiveness of the Pythagorean identity make it a valuable tool in mathematical proofs and problem-solving.

Solution Verification Process

Verifying that a function satisfies a given differential equation involves several logical steps. For the exercise at hand, the verification process starts with substituting the function into the equation. First, calculate the derivative:

For $ y = \sin(x + C) $, we derive $ \frac{dy}{dx} = \cos(x + C) $.
For a constant $ y = \pm 1 $, the derivative is $ 0 $.

Next, substitute these derivatives into the differential equation to check if it holds:

Substituting $ \cos(x + C) $ and $ \sin(x + C) $ in, we obtain $ (\cos(x + C))^2 + (\sin(x + C))^2 = 1 $, confirming the identity.
When $ y = \pm 1 $, derive $ 0^2 + 1^2 = 1 $.

These calculations verify that the given solutions satisfy the differential equation, completing the solution verification process.
This step-by-step verification helps ensure that the solutions are consistent with the original problem and offer insight into their validity.

Problem 4

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