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