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 ( yleft(\frac{pi}{2}
ight) ) for the function ( y = \frac{cos x}{x} ).
( yleft(\frac{pi}{2}
ight) = square ) (simplify your answer. type an integer or a simplified fraction )
is the function ( y = \frac{cos x}{x} ) a solution of ( xy + y = -sin x )? choose the correct answer below.
yes
no

Explanation:

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

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = \cos x\), \(u^\prime=-\sin x\), \(v = x\), \(v^\prime = 1\).
\(y^\prime=\frac{-\sin x\cdot x-\cos x\cdot1}{x^{2}}=\frac{-x\sin x - \cos x}{x^{2}}\)

Step2: Substitute \(y\) and \(y^\prime\) into \(xy^\prime + y\)

\(x\cdot\frac{-x\sin x-\cos x}{x^{2}}+\frac{\cos x}{x}\)
\(=\frac{-x\sin x-\cos x}{x}+\frac{\cos x}{x}\)
\(=\frac{-x\sin x-\cos x+\cos x}{x}\)
\(=\frac{-x\sin x}{x}=-\sin x\)

Answer:

Yes