Question

Show that \(y=\cos x, y=\sin x, y=c_{1} \cos x, y=c_{2} \sin x\) are all solutions of the differential equation \(\mathrm{y}_{2}+\mathrm{y}=0\)

Short answer

Question: Verify that the functions \(y = \cos x\), \(y = \sin x\), \(y = c_1 \cos x\), and \(y = c_2 \sin x\) are solutions of the differential equation \(y'' + y = 0\). Answer: After finding the second derivatives of each function and substituting them into the differential equation, we found that in each case, the equation is satisfied. Therefore, the functions \(y = \cos x\), \(y = \sin x\), \(y = c_1 \cos x\), and \(y = c_2 \sin x\) are solutions of the differential equation \(y'' + y = 0\).

Step-by-step solution

Find the second derivatives of the given functions

First, we need to find the second derivatives for each of the given functions: For \(y = \cos x\), \(y' = -\sin x\), and \(y'' = -\cos x\). For \(y = \sin x\), \(y' = \cos x\), and \(y'' = -\sin x\). For \(y = c_1 \cos x\), \(y' = -c_1 \sin x\), and \(y'' = -c_1 \cos x\). For \(y = c_2 \sin x\), \(y' = c_2 \cos x\), and \(y'' = -c_2 \sin x\).

Substitute the second derivatives into the equation

Now we substitute the second derivatives and the original functions into the differential equation \(y'' + y = 0\): For \(y = \cos x\), we have \((-\cos x) + (\cos x) = 0\). For \(y = \sin x\), we have \((-\sin x) + (\sin x) = 0\). For \(y = c_1 \cos x\), we have \((-c_1 \cos x) + (c_1 \cos x) = 0\). For \(y = c_2 \sin x\), we have \((-c_2 \sin x) + (c_2 \sin x) = 0\).

Confirm that the equation is satisfied

In each case, the left-hand side of the equation is equal to the right-hand side, which means that all given functions are solutions of the differential equation \(y'' + y = 0\). Therefore, we have shown that \(y = \cos x\), \(y = \sin x\), \(y = c_1 \cos x\), and \(y = c_2 \sin x\) are all solutions of the differential equation \(y'' + y = 0\).

Key concepts

Second Derivative

The second derivative is a measure of how the rate of change of a quantity is itself changing. In the context of the exercise, finding the second derivative is crucial for verifying solutions to differential equations. When dealing with functions like trigonometric ones, the second derivative helps identify the nature of their motion. For instance, a positive second derivative indicates that the graph of the function is concave up, representing an acceleration in the 'upward' direction, while a negative second derivative means concave down, denoting an acceleration 'downward'.

In our example, we computed the second derivatives of both cosine and sine functions, as well as their respective constants multiplied by them, and observed that they are simply the negatives of the original functions. This property is characteristic of harmonic functions and confirms that they satisfy the given differential equation \( y'' + y = 0 \).

Trigonometric Functions

Trigonometric functions such as sine (\( \text{sin} x \)) and cosine (\( \text{cos} x \)) are periodic functions that repeat their values in a predictable pattern. These functions are fundamental when solving differential equations involving oscillatory motion. Their derivatives and second derivatives play a key role in characterizing the behavior of systems described by such equations.

In the given exercise, \( \text{sin} x \) and \( \text{cos} x \) as solutions to the differential equation indicate that the motion represented by the equation is periodic, just like the trigonometric functions themselves. The fact that their second derivatives are negative multiples of the functions is what links them to the concept of harmonic motion.

Harmonic Motion

Harmonic motion refers to a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. This kind of motion is beautifully modeled by trigonometric functions. For an object in harmonic motion, such as a mass on a spring or a pendulum swinging, its position over time is often described by sine and cosine functions.

The exercise demonstrates how trigonometric functions solve the differential equation of harmonic motion (\( y'' + y = 0 \) by their nature. Mathematically, if you take a trigonometric function that models harmonic motion and find its second derivative, you will end up with the same function but with a negative sign. This reinforces the concept that in harmonic motion, acceleration (second derivative of position) is always directed opposite to displacement.

Simple Harmonic Oscillator

A simple harmonic oscillator is a system that experiences harmonic motion in the absence of damping or external forces. The classic examples include a mass on a spring or a simple pendulum. The behavior of a simple harmonic oscillator is described by the second-order linear differential equation \( y'' + ky = 0 \), where \( k \) is a constant that depends on the properties of the system, such as mass or spring constant.

The solutions to this differential equation, as shown in the textbook exercise, are the sine and cosine functions and any combination of these functions with constants, such as \( c_1 \) and \( c_2 \) in the exercise. These solutions imply that an undamped simple harmonic oscillator's motion can be fully described by these trigonometric functions, illustrating the timelessness and uniformity of such physical systems.