Answer
$y=x^{-3}(\sin x+c)$
Work Step by Step
The standard form of the given equation is:
$y'+\dfrac{3}{x} y=x^{-3} \cos x$ ....(1)
The integrating factor is: $e^{\int (\dfrac{3}{x}) dx}=x^3$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [x^3y]' =\int \cos x dx$
Thus, we have
$y=x^{-3}(\sin x+c)$
