\[ x y'(x)+y(x)+y(x)^2 (-\log (x))=0 \] ✓ Mathematica : cpu = 0.0744407 (sec), leaf count = 15
\[\left \{\left \{y(x)\to \frac {1}{\log (x)+c_1 x+1}\right \}\right \}\] ✓ Maple : cpu = 0.02 (sec), leaf count = 13
\[y \relax (x ) = \frac {1}{1+c_{1} x +\ln \relax (x )}\]
Hand solution
\(xy^{\prime }+axy^{2}+2y+bx=0\)This is Riccati non-linear first order. Converting it to standard form\begin {align} xy^{\prime }-y^{2}\ln x+y & =0\tag {1}\\ y^{\prime } & =-\frac {1}{x}y+y^{2}\frac {\ln x}{x}\nonumber \\ & =f_{0}+f_{1}y+f_{2}y^{2}\nonumber \end {align}
This is Bernoulli non-linear first order ODE since \(f_{0}=0\). Dividing by \(y^{2}\) gives
\[ \frac {y^{\prime }}{y^{2}}=-\frac {1}{x}\frac {1}{y}+\frac {\ln x}{x}\]
Let \(u=\frac {1}{y}\), hence \(u^{\prime }=-\frac {y^{\prime }}{y^{2}}\), and the above becomes
\begin {align*} -u^{\prime } & =-\frac {1}{x}u+\frac {\ln x}{x}\\ u^{\prime }-\frac {1}{x}u & =-\frac {\ln x}{x} \end {align*}
Integrating factor is \(\mu =e^{\int -\frac {1}{x}dx}=e^{-\ln x}=\frac {1}{x}\), hence
\[ d\left (\mu u\right ) =-\mu \frac {\ln x}{x}\]
Integrating
\begin {align*} \frac {1}{x}u & =-\int \frac {1}{x^{2}}\ln xdx+C\\ & =-\left (-\frac {\ln x}{x}-\frac {1}{x}\right ) +C \end {align*}
Therefore
\[ u=\ln x+1+Cx \]
Since \(u=\frac {1}{y}\) then
\[ y=\frac {1}{\ln x+1+Cx}\]
Verification