Answer
$y=\dfrac{1}{2}-\dfrac{1}{x}+\dfrac{c}{x^2}$
Work Step by Step
The standard form of the given equation is:
$y'+(\dfrac{2}{x})y=\dfrac{1}{x}-\dfrac{1}{x^2}$ ....(1)
The integrating factor is: $e^{\int (\dfrac{2}{x}) dx}=x^2$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [x^2y]' =\int (x-1) dx$ ...(2)
Thus, equation (2) becomes:
$y=\dfrac{1}{2}-\dfrac{1}{x}+\dfrac{c}{x^2}$
