Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 536: 10

$y=\dfrac{1}{x^2}(\sin x+c)$

Re-write the given differential equation as: $y'+\dfrac{2y}{x}=\dfrac{\cos x}{x^2}$ The integrating factor is: $e^{\int (2/x) dx}=x^2$ Now, we have $\int [x^2y]' dx=\int \cos x dx$ or, $x^2y=\sin x+c$ Hence, the General solution is: $y=\dfrac{1}{x^2}(\sin x+c)$