Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 11

$y=\dfrac{1}{2}-\dfrac{1}{x}+\dfrac{c}{x^2}$

Here, we have $y'+(\dfrac{2}{x})y=\dfrac{1}{x}-\dfrac{1}{x^2}$ The integrating factor is: $I=e^{\int (\dfrac{2}{x}) dx}=x^2$ Now, $x^2[y'+(\dfrac{2}{x})y]=x^2[\dfrac{1}{x}-\dfrac{1}{x^2}]$ This gives: $\int [x^2y]' =\int (x-1) dx$ Hence, $y=\dfrac{1}{2}-\dfrac{1}{x}+\dfrac{c}{x^2}$