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 four functions verify the differential equation \(y'' = -y\).

Step-by-step solution

Find the second derivative of each function

Begin by finding the second derivative of each function provided in the exercise.For \(y = \sin x\): The first derivative: \(y' = \cos x\)The second derivative: \(y'' = -\sin x\)For \(y = \cos x\): The first derivative: \(y' = -\sin x\)The second derivative: \(y'' = -\cos x\)For \(y = e^{i x}\): The first derivative: \(y' = i e^{i x}\)The second derivative: \(y'' = i^2 e^{i x} = -e^{i x}\)For \(y = e^{-i x}\): The first derivative: \(y' = -i e^{-i x}\)The second derivative: \(y'' = (-i)^2 e^{-i x} = -e^{-i x}\)

Verify the relationship \(y'' = -y\) for each function

Now, check if the second derivative of each function equals \(-y\).For \(y = \sin x\): The second derivative is \(y'' = -\sin x = -y\)For \(y = \cos x\): The second derivative is \(y'' = -\cos x = -y\)For \(y = e^{i x}\): The second derivative is \(y'' = -e^{i x} = -y\)For \(y = e^{-i x}\): The second derivative is \(y'' = -e^{-i x} = -y\)

Conclusion

Since the second derivative of each function is equal to the negative of the original function, \(-y\), all four functions are indeed solutions to the differential equation \(y'' = -y\).

Key concepts

second derivative

The second derivative of a function measures the rate at which the first derivative changes. It provides key insights into the concavity and inflection points of the function.

For the exercise given, let's revisit how to find the second derivatives:

• For the function \(y = \sin x\), the first derivative is \(y' = \cos x\) and the second derivative is \(y'' = -\sin x\).
• For the function \(y = \cos x\), the first derivative is \(y' = -\sin x\) and the second derivative is \(y'' = -\cos x\).
• For the function \(y = e^{i x}\), the first derivative is \(y' = i e^{i x}\) and the second derivative is \(y'' = i^2 e^{i x} = -e^{i x}\).
• For the function \(y = e^{-i x}\), the first derivative is \(y' = -i e^{-i x}\) and the second derivative is \(y'' = (-i)^2 e^{-i x} = -e^{-i x}\).

The second derivative helps in verifying solutions to certain differential equations. In this case, each second derivative equaled \(-y\) confirming they are solutions of the differential equation \(y'' = -y\).

trigonometric functions

Trigonometric functions such as sine and cosine are essential in studying oscillatory behaviors.

In the given exercise:

• The function \(y = \sin x\) has its first derivative \(y' = \cos x\) and its second derivative \(y'' = -\sin x\), aligning with \(y'' = -y\).
• Similarly, for \(y = \cos x\), the first derivative is \(y' = -\sin x\) and the second derivative is \(y'' = -\cos x\), also fitting the equation \(y'' = -y\).

Trigonometric functions are periodic, repeating values in regular intervals. This property often simplifies solving differential equations involving these functions.

complex exponentials

Complex exponentials, involving the imaginary unit \(i\), often simplify computations in oscillatory systems. The Euler's formula states, \(e^{ix} = \cos x + i\sin x\).

In the exercise:

• For \(y = e^{i x}\), the derivatives are \(y' = i e^{i x}\) and \(y'' = i^2 e^{i x} = -e^{i x}\).
• Similarly, for \(y = e^{-i x}\), the derivatives are \(y' = -i e^{-i x}\) and \(y'' = (-i)^2 e^{-i x} = -e^{-i x}\).

These results confirm that both functions fit the equation \(y'' = -y\).

Complex exponentials are crucial in physics and engineering, especially in solving problems involving waves and harmonic oscillations.