Question

Determining a Solution In Exercises \(13-20\) , determine whether the function is a solution of the differential equation \(y^{(4)}-16 y=0\) . $$ y=3 \sin 2 x $$

Short answer

Yes, \(y=3 \sin 2x\) is a solution of the differential equation \(y^{(4)}-16 y=0\)

Step-by-step solution

Identifying the given function and the differential equation

The given function is \(y=3 \sin 2x\) and the differential equation is \(y^{(4)}-16 y=0\)

Calculating the first derivative of y

To calculate the derivative of \(y=3 \sin 2x\), apply the chain rule which gives the first derivative as \(y' = 6 \cos 2x\)

Calculating the second derivative of y

The second derivative, \(y''\), of \(y=3 \sin 2x\) is \(-12 \sin 2x\) as the derivative of a cosine function is negative sine function

Calculating the third derivative of y

Again applying the derivative rules, the third derivative, \(y'''\), of \(y=3 \sin 2x\) is \(-24 \cos 2x\) because the derivative of a sine function is a cosine function

Calculating the fourth derivative of y

As the derivative of a cosine function is negative sine, the fourth derivative, \(y^{(4)}\), of \(y=3 \sin 2x\) is \(48 \sin 2x\) using the chain rule

Substituting the function and its fourth derivative into the differential equation

Substituting \(y=3 \sin 2x\) and \(y^{(4)}=48 \sin 2x\) in \(y^{(4)}-16y=0\), gives \(48 \sin 2x-16(3 \sin 2x)\)

Checking if the equation holds true

Simplifying the equation obtained in Step 6 gives \(0\). On putting the values of y and \(y^{(4)}\), the left hand side equals right hand side, thus confirming that \(y=3 \sin 2x\) is a solution of the given differential equation

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to compute the derivative of a composition of two or more functions. If you have a function of a function, like our exercise here with \( y = 3 \sin(2x) \), the chain rule is the tool that helps you find the derivative efficiently.

Imagine the process of peeling an onion—it's layer by layer. Similarly, when applying the chain rule in calculus, you differentiate the outer function, leaving the inner function unchanged (like the first layer), then multiply by the derivative of the inner function (the next layer). For the given function \( y = 3 \sin(2x) \), when we take its first derivative, we treat \( \sin(2x) \), which is the inner function, as a single unit first. So we have \( \frac{d}{dx}3 \sin(2x) = 3 \frac{d}{dx}\sin(2x) \). Then using the chain rule, we multiply the derivative of this unit by the derivative of its inside part, which is \( 2 \), giving us \( 6 \cos(2x) \).

This concept is crucial in our exercise as we apply it repetitively to find higher order derivatives like the second \( (y'') \), third \( (y''') \), and the fourth derivative \( (y^{(4)}) \) for the equation given.

Derivatives

In the realm of calculus, derivatives represent the rate at which a function is changing at any given point. They are a cornerstone concept that allows us to understand movement, growth, and change across different disciplines like physics, economics, and biology.

More technically, the derivative of a function at a point is the slope of the tangent line to the function's graph at that point. It is represented as \( f'(x) \) or \( \frac{df}{dx} \). In our exercise, we begin with the function \( y = 3 \sin(2x) \) and take its derivative to measure how \( y \) changes as \( x \) changes. Derivatives come in orders - the first derivative \( y' \) is the basic rate of change, while higher order derivatives, like the second, third, and fourth we calculate, describe more complex types of change such as acceleration.

Understanding how to find derivatives is crucial because it allows us to predict and understand patterns in motion and change when we apply them to real-world situations.

Fourth Derivative

While the first derivative gives the rate of change, and the second derivative provides the rate of change of the rate of change (often called acceleration), the fourth derivative is not as commonly interpreted but is essential in certain fields such as physics and engineering.

In our problem, we're concerned with the fourth derivative \( y^{(4)} \) of the function \( y = 3 \sin(2x) \). Each differentiation step peels back another layer, revealing the function's deeper changes. By applying the chain rule, as well as the derivatives of sin and cosine functions (addressed in the next section), we repetitively derive to reach the fourth derivative. As shown in the solution steps, for \( y=3 \sin 2x \), the fourth derivative is \( 48 \sin 2x \).

Finding the fourth derivative can sometimes help in analyzing the behavior of the original function, like whether it's experiencing repeated oscillating motion, which is common in systems like springs or waves, hence its importance in physics and other disciplines.

Sin and Cosine Functions

The functions sine (\( \sin(x) \) ) and cosine (\( \cos(x) \) ) are two of the most important trigonometric functions, and they play a central role when we deal with periodic phenomena such as waves or circular motion.

\clearpage The sine function represents the y-coordinate of a point on a unit circle at an angle \( x \) from the positive x-axis, while the cosine function gives the x-coordinate. They have a unique relationship and are foundational in both pure and applied mathematics. For instance, in our exercise, when we differentiate \( y = 3 \sin(2x) \) to get \( y' = 6 \cos(2x) \), we're using the fact that the derivative of \( \sin \) is \( \cos \) and vice versa, but with a change in sign depending on the order of the derivative.

Moreover, these functions have periodic properties, meaning they repeat their values in regular intervals, making them suitable for describing oscillatory motions. In the context of the exercise, we use the derivative properties of these functions to solve the differential equation involving the fourth derivative of a sinusoidal function.