Answer
$y= \frac{1}{x}e^{\frac{C}{x^2}}$
Work Step by Step
We are given
$xy[2\ln xy+1]dx+x^2dy=0$
Here, $M(x,y)=xy[2\ln xy+1]$
$N(x,y)=x^2$
We have:
$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{N}=\frac{-1}{y}$
The integrating factor is:
$I(x)=e^{\int f(x)dx}=e^{\int \frac{-1}{y}dy}=e^{-\ln y}=\frac{1}{y}$
Multiplying equation (1) by $I(x)=\frac{1}{y}$
$\frac{1}{y}(xy(2\ln xy+1)dx+x^2dy)=0$
$x(2\ln xy+1)dx+(x^2y^{-1})dy=0$
Then:
$\frac{\partial \phi}{\partial x}=x(2\ln xy+1)$
Integrating both sides:
$\phi=x^2+\ln (xy)+f(y)$
We have:
$\frac{\partial phi}{\partial y}=\frac{x^2}{y}+f'(y)=\frac{x^2}{y}$
Since $f'(y)=0 \rightarrow f(y)=C$
C is a constant integration
The solution is:
$\phi (x,y)=x^2\ln xy+C$
or
$y= \frac{1}{x}e^{\frac{C}{x^2}}$
