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 ).
find ( xy ) for the function ( y = \frac{cos 3x}{x} ).
( xy = )

Explanation:

Step1: Differentiate \( y = \frac{\cos3x}{x} \) using quotient rule

The quotient rule is \( (u/v)'=\frac{u'v - uv'}{v^{2}} \), where \( u = \cos3x \), \( u'=- 3\sin3x \), \( v = x \), \( v' = 1 \).
\( y'=\frac{-3x\sin3x-\cos3x}{x^{2}} \)

Step2: Calculate \( xy' \)

Multiply \( y' \) by \( x \):
\( xy'=\frac{-3x\sin3x-\cos3x}{x}=-3\sin3x-\frac{\cos3x}{x} \)

Answer:

\(-3\sin3x-\frac{\cos3x}{x}\)