Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 12

$y=x [\ln x]^2 +cx$

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