Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 16

$y=x^{-3}(\sin x+c)$

Here, we have $y'+\dfrac{3}{x} y=x^{-3} \cos x$ The integrating factor is: $I=e^{\int (\dfrac{3}{x}) dx}=x^3$ Now, $x^3[y'+\dfrac{3}{x} y]=(x^3)(x^{-3}) (\cos x)$ This implies that $\int [x^3y]' =\int \cos x dx$ Hence, $y=x^{-3}(\sin x+c)$