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Chapter 8: Problem 3

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

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

Step by step solution

01

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: \(y'' = (-i)^2 e^{-i x} = -e^{-i x}\)
02

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 \(y'' = -e^{-i x} = -y\)
03

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 \(y'' = -y\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is \(2 \pi \sqrt{h / g}\) where \(h\) is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.

By separation of variables, find a solution of the equation \(y^{\prime}=\sqrt{y}\) containing one arbitrary constant. Find a particular solution satisfying \(y=0\) when \(x=0 .\) Show that \(y \equiv 0\) is a solution of the differential equation which cannot be obtained by specializing the arbitrary constant in your solution above. Computer plot a slope field and some of the solution curves. Show that there are an infinite number of solution curves passing through any point on the \(x\) axis, but just one through any point for which \(y>0 .\) Hint: See Example 3. Problems 17 and 18 are physical problems leading to this differential equation.

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-5 y^{\prime}+6 y=e^{2 x}$$

Find a particular solution satisfying the given conditions. \(3 x^{2} y d x+x^{3} d y=0, \quad y=2\) when \(x=1\).

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(2 x-y \sin 2 x) d x=\left(\sin ^{2} x-2 y\right) d y$$
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