Question

check whether the function ( y = \frac{cos 3x}{x} ) is a solution of ( xy + y = - 3 sin 3x ) with the initial condition ( y left( \frac{pi}{2}
ight) = 0 ).
find ( xy ) for the function ( y = \frac{cos 3x}{x} ).
( xy = - 3 sin 3x - \frac{cos 3x}{x} )
find ( xy + y ) for the function ( y = \frac{cos 3x}{x} ).
( xy + y = - 3 sin 3x )
find ( y left( \frac{pi}{2}
ight) ) for the function ( y = \frac{cos 3x}{x} ).
( y left( \frac{pi}{2}
ight) = square ) (simplify your answer. type an integer or a simplified fraction )

Explanation:

Step1: Substitute \(x = \frac{\pi}{2}\) into \(y=\frac{\cos3x}{x}\)

$$y(\frac{\pi}{2})=\frac{\cos(3\times\frac{\pi}{2})}{\frac{\pi}{2}}$$

Step2: Calculate \(\cos(3\times\frac{\pi}{2})\)

We know that \(\cos(\frac{3\pi}{2}) = 0\)

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

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

Answer:

\(0\)