Question

Verify that \(y=\sin x, y=\cos x, y=e^{i x}\), and \(y=e^{-i x}\) are all solutions of \(y^{\prime \prime}=-y\).

Short answer

All functions, \(y = \sin x, y = \cos x, y = e^{i x}, \) and \(y = e^{-i x}\), satisfy the differential equation \(y'' = -y\).

Step-by-step solution

Differentiate the Function

Find the second derivative of each function to check if it satisfies the given differential equation, \(y'' = -y\).

Check y = sin(x)

The first derivative of \(y = \sin x\) is \(y' = \cos x.\) The second derivative is \(y'' = -\sin x.\) Substituting into the differential equation \(y'' = -y\): \(-\sin x = -\sin x\). This holds true.

Check y = cos(x)

The first derivative of \(y = \cos x\) is \(y' = -\sin x.\) The second derivative is \(y'' = -\cos x.\) Substituting into the differential equation \(y'' = -y\): \(-\cos x = -\cos x\). This holds true.

Check y = e^{ix}

The first derivative of \(y = e^{i x}\) is \(y' = i e^{i x}.\) The second derivative is \(y'' = i^2 e^{i x} = -e^{i x}.\) Substituting into the differential equation \(y'' = -y\): \(-e^{i x} = -e^{i x}\). This holds true.

Check y = e^{-ix}

The first derivative of \(y = e^{-i x}\) is \(y' = -i e^{-i x}.\) The second derivative is \(y'' = (-i)^2 e^{-i x} = -e^{-i x}.\) Substituting into the differential equation \(y'' = -y\): \(-e^{-i x} = -e^{-i x}\). This holds true.

Key concepts

Harmonic Functions

Harmonic functions, such as sine and cosine, are solutions to certain types of differential equations. A function is called harmonic if it satisfies the Laplace equation. In this context, we are dealing with the simpler form of harmonic functions that satisfy a specific second-order differential equation:
For example, in the exercise given, both \(y = \sin x\) and \(y = \cos x\) are solutions to the second-order differential equation \(y'' = -y\). The term 'harmonic' comes from these functions' property of producing smooth, wave-like oscillations, which makes sine and cosine fundamental in the study of oscillatory phenomena such as sound and light waves. These harmonic functions are periodic, with \(\sin x\) and \(\cos x\) having a period of \(2 \pi \).

Complex Exponentials

Complex exponentials take the form \(e^{ix}\) where \(i\) is the imaginary unit. Complex exponentials are deeply tied to trigonometric functions through Euler's formula: \(e^{ix} = \cos x + i\sin x\). In the given exercise, both \(e^{ix}\) and \(e^{-ix}\) are solutions for the differential equation \(y'' = -y\). To see why, differentiate \(e^{ix}\): its first derivative is \(ie^{ix}\), and its second derivative is \(i^2e^{ix} = -e^{ix}\), fitting our differential equation perfectly. Complex exponentials are key in representing oscillations and waves, especially in physics and engineering.

Differentiation

Differentiation is a fundamental concept in calculus, used to find the rate at which a function is changing at any given point. For second-order differential equations like \(y'' = -y\), we need to differentiate a function twice. Let's look at this step-by-step:
1. **First Derivative**: For \(y = \sin x\), the first derivative is \(y' = \cos x\). This tells us the rate of change of \(\sin x\) at any point.

2. **Second Derivative**: Differentiating \(\cos x\) gives us \(y'' = -\sin x\). This second derivative expresses how the rate of change of \(\cos x\) itself is changing.
Differentiation is thus crucial for solving and verifying differential equations, like we did in this exercise. It allows us to analyze the behavior of functions in depth and verify that they satisfy the required conditions.