Question

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

(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

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) \(y = \sin x\)

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.

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.

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.

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.