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Chapter 4: Problem 2

Are the following functions solutions of the equation \(y^{\prime}+y \cos x=\frac{1}{2} \sin 2 x ?\) (a) \(\mathrm{y}=\sin \mathrm{x}-1\) (b) \(y=e^{-\sin x}\) (c) \(y=\sin x\)

Short Answer

Expert verified
(a) \(y = \sin x - 1\) (b) \(y = e^{-\sin x}\) (c) \(y = \sin x\) Answer: (b) \(y = e^{-\sin x}\) and (c) \(y = \sin x\) are solutions to the given equation.

Step by step solution

01

Find the first derivative of the given functions

Using the definition of derivatives, find the first derivative of each function: (a) \(y = \sin x - 1\) (b) \(y = e^{-\sin x}\) (c) \(y = \sin x\)
02

Substitute each function and its derivative in the given equation

Substitute the original function and its derivative into the given equation \(y^{\prime} + y\cos x = \frac{1}{2}\sin 2x\) and check if it holds true.
03

Test the functions (a) \(y = \sin x - 1\)

First, find the derivative: \(y^\prime = \cos x\) Substitute \(y\) and \(y^\prime\) into the equation: \((\cos x) + (\sin x - 1)\cdot \cos x = \frac{1}{2}\cdot \sin(2x)\) This simplifies to: \(\cos x + \sin x \cdot \cos x - \cos^2 x = \frac{1}{2}\cdot\sin(2x)\) This equation does not hold true, so the function (a) is not a solution.
04

Test the functions (b) \(y = e^{-\sin x}\)

First, find the derivative using the chain rule: \(y^\prime = -\cos x \cdot e^{-\sin x}\) Substitute \(y\) and \(y^\prime\) into the equation: \((-\cos x \cdot e^{-\sin x}) + (e^{-\sin x})\cdot \cos x = \frac{1}{2}\cdot \sin(2x)\) This simplifies to: \(-\cos x \cdot e^{-\sin x} + \cos x\cdot e^{-\sin x} = 0\) Both sides of the equation are zero, so the function (b) is a solution.
05

Test the functions (c) \(y = \sin x\)

First, find the derivative: \(y^\prime = \cos x\) Substitute \(y\) and \(y^\prime\) into the equation: \((\cos x) + (\sin x) \cdot \cos x = \frac{1}{2}\cdot \sin(2x)\) This simplifies to: \(\cos x + \sin x \cdot \cos x = \frac{1}{2}\cdot \sin(2x)\) This equation holds true, so the function (c) is a solution. To summarize, functions (b) \(y = e^{-\sin x}\) and (c) \(y = \sin x\) are solutions of the given equation, while function (a) \(y = \sin x-1\) is not a solution.

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