#### Answer

$$y = \frac{{{x^2}}}{{\ln x}} + \frac{C}{{\ln x}}$$

#### Work Step by Step

$$\eqalign{
& x\ln x\frac{{dy}}{{dx}} + y = 2{x^2} \cr
& {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{divide both sides of the equation by }}x\ln x \cr
& \frac{{dy}}{{dx}} + \frac{1}{{x\ln x}}y = \frac{{2x}}{{\ln x}} \cr
& {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{ we can note that }}P\left( x \right){\text{ is }}\frac{1}{{x\ln x}} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{1}{{x\ln x}}} dx}} = {e^{\ln \left| {\ln x} \right|}} = \ln x \cr
& {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} + \frac{1}{{x\ln x}}y = \frac{{2x}}{{\ln x}}{\text{ by }}\ln x \cr
& \ln x\frac{{dy}}{{dx}} + \frac{1}{x}y = 2x \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {y\ln x} \right] = 2x \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& y\ln x = \int {2x} dx \cr
& y\ln x = {x^2} + C \cr
& y = \frac{{{x^2}}}{{\ln x}} + \frac{C}{{\ln x}} \cr} $$