#### Answer

$y=x(\ln x)^2+xc$

#### Work Step by Step

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