Problem 28 - Calculus

Question

If \(y=\sin 5 x\) then \(y^{\prime}=5 \cos 5 x\) and \(y^{\prime \prime}=-25 \sin 5 x .\) So this function satisfies the differential equation \(y^{\prime \prime}=\) ______ .

Step-by-Step Solution

Answer

The differential equation is \(y'' = -25y\).

Step 1: Understand the Given Function

The function given is \(y = \sin 5x\). This represents a basic trigonometric function that is reliant on \(x\), with the inner function multiplied by 5. We need to find its higher derivatives to see which differential equation it satisfies.

Step 2: Compute the First Derivative

To find \(y'\), the first derivative of \(y\), we apply the chain rule. The derivative of \(\sin u\) is \(\cos u\), then multiply by the derivative of \(u\) (which is \(5x\) in this case). Thus, \(y' = 5\cos 5x\).

Step 3: Compute the Second Derivative

Next, to find \(y''\), take the derivative of \(y' = 5 \cos 5x\). Again, using the chain rule, the derivative of \(\cos u\) is \(-\sin u\), and multiply it by the derivative of \(u\) (\(5x\)) gives: \(y'' = 5(-5\sin 5x) = -25 \sin 5x\).

Step 4: Write the Differential Equation

We found from step 3 that \(y'' = -25 \sin 5x\). This shows the relationship between \(y''\) and the function \(y\). Therefore, the differential equation it satisfies can be expressed as \(y'' = -25y\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in many areas of mathematics and science. They relate the angles of a triangle to the lengths of its sides. The most commonly used trigonometric functions are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). In our exercise, we focus on the sine function, specifically \( y = \sin 5x \), which represents a periodic waveform.

These functions are essential because:

• They help describe and model periodic phenomena such as sound and light waves.
• They are key in solving problems involving right-angled triangles and circular motion.
• Each trigonometric function has a specific wave pattern that can be used to describe oscillations or cycles.

In this exercise, \( 5x \) is the argument of the sine function, indicating a horizontal stretch or compression by a factor of 5. Understanding how this affects the graph is key to solving differential equations involving trigonometric functions.

Derivatives

Derivatives represent the rate of change of a function concerning one of its variables. Finding the derivative of a function allows you to understand how the function behaves. This concept is crucial in calculus.

For our exercise, finding the derivatives of the trigonometric function \( y = \sin 5x \) helps us understand its rate of change.

• First Derivative (\( y' \)): This represents the slope of the function. In our example, the first derivative is shown as \( y' = 5 \cos 5x \), where you apply the chain rule.
• Second Derivative (\( y'' \)): This offers insights into the concavity of the function. The second derivative for \( y = \sin 5x \) is \( y'' = -25 \sin 5x \).

Derivatives are useful for understanding trends, optimizing problems, and solving differential equations, providing more profound insights into physical problems.

Chain Rule

The chain rule is a critical tool in calculus for differentiating functions that are compositions of two or more other functions. It essentially tells us how to take the derivative of a composite function.

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

• In our exercise, \( y = \sin 5x \)is a composite function, where the external function is \( \sin u \)and \( u = 5x \).
• The chain rule helps us find \( y' = 5\cos 5x \), indicating that \( \sin \)is differentiated to \( \cos \)and \( 5x \)is multiplied by its derivative, which is 5.

Using the chain rule allows us to simplify and solve problems involving complex functions. It is particularly useful in engineering and physics where multiple variables often interact.