Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 536: 3

$y=\dfrac{-\cos x}{x^3}+\dfrac{c}{x^3}$ or, $y=\dfrac{-\cos x+c}{x^3}$

Re-write the given differential equation as: $x^3y+3y=\sin x$ or, $(x^3y)'=\sin x$ Now, we have $\int (x^3y)' dx=\int \sin x dx$ Then $x^3 y=-\cos x+c$ Hence, the General solution is: $y=\dfrac{-\cos x}{x^3}+\dfrac{c}{x^3}$ or, $y=\dfrac{-\cos x+c}{x^3}$