Question

check whether the function ( y = \frac{cos 3x}{x} ) is a solution of ( xy + y = -3sin 3x ) with the initial condition ( yleft( \frac{pi}{2}
ight) = 0 ).

( xy + y = -3sin 3x )

find ( yleft( \frac{pi}{2}
ight) ) for the function ( y = \frac{cos 3x}{x} ).

( yleft( \frac{pi}{2}
ight) = 0 ) (simplify your answer. type an integer or a simplified fraction.)

is the function ( y = \frac{cos 3x}{x} ) a solution of ( xy + y = -3sin 3x )? choose the correct answer below.

yes

no

Explanation:

Step1: Find the derivative of \(y = \frac{\cos3x}{x}\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = \cos3x\), \(u^\prime=- 3\sin3x\), \(v = x\), \(v^\prime = 1\).

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Step2: Calculate \(xy^\prime + y\)

Substitute \(y=\frac{\cos3x}{x}\) and \(y^\prime=\frac{-3x\sin3x - \cos3x}{x^{2}}\) into \(xy^\prime + y\).

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Answer:

A. Yes