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