\[ \frac {dy}{dx}=\frac {2y^{2}-xy}{3xy-2x^{2}}\] Let \(y=ux\) or \(u=\frac {y}{x}\), hence \(\frac {dy}{dx}=x\frac {du}{dx}+u\) and the above ode becomes\begin {align*} x\frac {du}{dx}+u & =\frac {2u^{2}x^{2}-x^{2}u}{3x^{2}u-2x^{2}}\\ x\frac {du}{dx}+u & =\frac {2u^{2}-u}{3u-2}\\ x\frac {du}{dx} & =\frac {2u^{2}-u}{3u-2}-u\\ & =\frac {2u^{2}-u}{3u-2}-\frac {u\left ( 3u-2\right ) }{3u-2}\\ & =\frac {\left ( 2u^{2}-u\right ) -u\left ( 3u-2\right ) }{3u-2}\\ & =\frac {2u^{2}-u-3u^{2}+2u}{3u-2}\\ & =\frac {-u^{2}+u}{3u-2}\\ & =\frac {u\left ( 1-u\right ) }{3u-2} \end {align*}
Hence\[ \frac {du}{dx}=\left ( \frac {1}{x}\right ) \left ( \frac {u\left ( 1-u\right ) }{3u-2}\right ) \] Which is separable. If we do not obtain separable ode, then we have made mistake. Integrating gives\begin {align*} \int \frac {3u-2}{u\left ( 1-u\right ) }du & =\int \frac {1}{x}dx\qquad \frac {u\left ( 1-u\right ) }{3u-2}\neq 0\\ -2\ln u-\ln \left ( u-1\right ) & =\ln x+c_{1} \end {align*}
Replacing \(u=\frac {y}{x}\) gives\begin {align*} -2\ln \left ( \frac {y}{x}\right ) -\ln \left ( \frac {y}{x}-1\right ) & =\ln x+c_{1}\\ \ln \left ( \frac {x^{2}}{y^{2}}\right ) -\ln \left ( \frac {y-x}{x}\right ) & =\ln x+c_{1}\\ \ln \left ( \frac {x^{2}}{y^{2}}\right ) +\ln \left ( \frac {x}{y-x}\right ) & =\ln x+c_{1} \end {align*}
Applying exponential to each side gives\begin {equation} \left ( \frac {x^{2}}{y^{2}}\right ) \left ( \frac {x}{y-x}\right ) =c_{2}x \tag {1} \end {equation}
Singular solution is when \(\frac {u\left ( 1-u\right ) }{3u-2}=0\). This gives \(u=0\) and \(u=1\). Hence this implies \(y=0\) and \(y=x\). Therefore the solutions are
\begin {align*} \left ( \frac {x^{2}}{y^{2}}\right ) \left ( \frac {x}{y-x}\right ) & =c_{2}x\\ y & =0\\ y & =x \end {align*}
Lets say that we had also initial conditions \(y\left ( 1\right ) =-1\), then the above gives\begin {align*} \left ( \frac {1}{-1-1}\right ) & =c_{2}\\ -\frac {1}{2} & =c_{2} \end {align*}
Therefore the solution (1) becomes\[ \left ( \frac {x^{2}}{y^{2}}\right ) \left ( \frac {x}{y-x}\right ) =-\frac {1}{2}x \]