Problem 4

Question

Show that if $B$ and $\omega$ are constants and $y(t)=B \cos (\omega t),$ then $$ y(0)=B \quad y^{\prime}(0)=0 \quad \text { and } \quad y^{\prime \prime}(t)+\omega^{2} y(t)=0 . $$

Step-by-Step Solution

Verified

Answer

The conditions are satisfied: $y(0) = B$, $y'(0) = 0$, and $y''(t) + \omega^2 y(t) = 0$.

1Step 1: Evaluate y(0)

To evaluate $y(0)$, substitute $t = 0$ into the given function $y(t) = B \cos(\omega t)$. This yields $y(0) = B \cos(0) = B$. Hence, $y(0) = B$.

2Step 2: Differentiate y(t) to find y'(t)

Differentiate the function $y(t) = B \cos(\omega t)$ with respect to $t$. The derivative is $y'(t) = -B \omega \sin(\omega t)$.

3Step 3: Evaluate y'(0)

Substitute $t = 0$ into the expression for $y'(t)$: $y'(0) = -B \omega \sin(0) = 0$. Thus, $y'(0) = 0$.

4Step 4: Differentiate y'(t) to find y''(t)

Differentiate $y'(t) = -B \omega \sin(\omega t)$ to find $y''(t)$. The derivative is $y''(t) = -B \omega^2 \cos(\omega t)$.

5Step 5: Verify the given differential equation

Substitute $y(t)$ and $y''(t)$ into the differential equation $y''(t) + \omega^2 y(t) = 0$: $-B \omega^2 \cos(\omega t) + \omega^2 (B \cos(\omega t)) = 0$, which simplifies to $0 = 0$. Hence, the differential equation holds true.

Key Concepts

Trigonometric FunctionsCalculusDerivative

Trigonometric Functions

Trigonometric functions are fundamental to understanding oscillations and waves, as they provide the means to describe periodic phenomena. In this problem, we deal with the cosine function represented as $ y(t) = B \cos(\omega t) $, where $ B $ is the amplitude, and $ \omega $ is the angular frequency.

These functions have specific properties that are particularly useful in solving differential equations. For example:

The cosine function is periodic, meaning it repeats after a certain interval. Its period is $ \frac{2\pi}{\omega} $, which tells us how fast the wave repeats in a cycle.
At $ t = 0 $, $ \cos(0) = 1 $, hence $ y(0) = B $.

Understanding the behavior of trigonometric functions at key points allows us to evaluate their initial conditions, such as $ y(0) $, which is crucial for solving differential equations involving these functions.

Calculus

Calculus is the mathematical study of continuous change, and it's the foundation behind analyzing rates of change and areas under curves. In problems involving functions like $ y(t) = B \cos(\omega t) $, calculus provides the tools to find derivatives and integrations, which help in understanding how functions behave over time.

Differentiation is a core concept of calculus used here to find the first derivative $ y'(t) $, which describes the rate of change of $ y(t) $ with respect to $ t $. We applied the chain rule:

The derivative of $ y(t) = B \cos(\omega t) $ leads us to $ y'(t) = -B \omega \sin(\omega t) $.

Similarly, further differentiation helps us find the acceleration or the second derivative $ y''(t) $, which is important for checking the validity of differential equations.

Calculus not only helps in finding derivatives but also assists in examining their applications, like verifying if a function satisfies a differential equation. Here, it leads us to check $ y''(t) + \omega^2 y(t) = 0 $ is indeed true.

Derivative

Derivatives play a vital role in understanding how functions change and react to different inputs. A derivative represents a function's instantaneous rate of change, similar to how speed describes how fast an object moves at a given moment.

In our problem, after differentiating $ y(t) = B \cos(\omega t) $, we found the first derivative $ y'(t) = -B \omega \sin(\omega t) $, indicating how quickly the function is changing at any point $ t $.

The derivative $ y'(t) = 0 $ at $ t = 0 $ shows there is no change in the function at this initial point.

Another level of differentiation provides the second derivative $ y''(t) = -B \omega^2 \cos(\omega t) $, which represents how the rate of change itself is changing.

It helps validate the original differential equation $ y''(t) + \omega^2 y(t) = 0 $ by substituting back the derivatives and confirming that both sides equate to zero.

Understanding derivatives is crucial to solving differential equations, evaluating rates of change, and exploring deeper mathematical relationships.

Problem 3

Problem 4

Other exercises in this chapter

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Problem 4

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