Question

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

Explanation:

Step1: Calculate \(y'\)

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

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Then \(xy'=x\cdot\frac{-x\sin x - \cos x}{x^{2}}=\frac{-x\sin x-\cos x}{x}=-\sin x-\frac{\cos x}{x}\)

Step2: Calculate \(xy' + y\)

Substitute \(y=\frac{\cos x}{x}\) and \(xy'=-\sin x-\frac{\cos x}{x}\) into \(xy' + y\)

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

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

$$y(\frac{\pi}{2})=\frac{\cos\frac{\pi}{2}}{\frac{\pi}{2}}=\frac{0}{\frac{\pi}{2}} = 0$$

Answer:

\(xy'=-\sin x-\frac{\cos x}{x}\)
\(xy' + y=-\sin x\)
\(y(\frac{\pi}{2}) = 0\)