Question

check whether the function ( y = \frac{cos 5x}{x} ) is a solution of ( xy + y = -5 sin 5x ) with the initial condition ( y(\frac{pi}{2}) = 0 ).
find ( xy ) for the function ( y = \frac{cos 5x}{x} ).
( xy = square )
find ( xy + y ) for the function ( y = \frac{cos 5x}{x} ).
( xy + y = square )
find ( y(\frac{pi}{2}) ) for the function ( y = \frac{cos 5x}{x} ).
( y(\frac{pi}{2}) = square ) (simplify your answer. type an integer or a simplified fraction)

Explanation:

Step1: Find the derivative \(y'\)

Use the quotient rule \((\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}\), where \(u = \cos5x\), \(u'=- 5\sin5x\), \(v = x\), \(v' = 1\).

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Then \(xy'=\frac{-5x\sin5x-\cos5x}{x}=-5\sin5x-\frac{\cos5x}{x}\)

Step2: Find \(xy' + y\)

Substitute \(xy'\) and \(y=\frac{\cos5x}{x}\) into \(xy' + y\)

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Step3: Find \(y(\frac{\pi}{2})\)

Substitute \(x = \frac{\pi}{2}\) into \(y=\frac{\cos5x}{x}\)

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Answer:

\(xy'=-5\sin5x-\frac{\cos5x}{x}\)
\(xy' + y=-5\sin5x\)
\(y(\frac{\pi}{2}) = 0\)