3.3.7.2 Example 2

$$\frac{dy}{dx}=\frac{2y^2-xy}{3xy-2x^2}$$

Let $y=ux$ or $u=\frac{y}{x}$, hence $\frac{dy}{dx}=x\frac{du}{dx}+u$ and the above ode becomes

$$\begin{array}{rl}x\frac{du}{dx}+u& =\frac{2u^2{x}^2-x^{2}u}{3x^{2}u-2x^2}\\ x\frac{du}{dx}+u& =\frac{2u^2-u}{3u-2}\\ x\frac{du}{dx}& =\frac{2u^2-u}{3u-2}-u\\ & =\frac{2u^2-u}{3u-2}-\frac{u(3u-2)}{3u-2}\\ & =\frac{(2u^2-u)-u(3u-2)}{3u-2}\\ & =\frac{2u^2-u-3u^2+2u}{3u-2}\\ & =\frac{-u^2+u}{3u-2}\\ & =\frac{u(1-u)}{3u-2}\end{array}$$

Hence

$$\frac{du}{dx}=\left(\frac{1}{x}\right)\left(\frac{u(1-u)}{3u-2}\right)$$

Which is separable. If we do not obtain separable ode, then we have made mistake. Integrating gives

$$\begin{array}{rl}\int \frac{3u-2}{u(1-u)}du& =\int \frac{1}{x}dx\frac{u(1-u)}{3u-2}\ne 0\\ -2\ln u-\ln (u-1)& =\ln x+c_1\end{array}$$

Replacing $u=\frac{y}{x}$ gives

$$\begin{array}{rl}-2\ln \left(\frac{y}{x}\right)-\ln (\frac{y}{x}-1)& =\ln x+c_1\\ \ln \left(\frac{x^2}{y^2}\right)-\ln \left(\frac{y-x}{x}\right)& =\ln x+c_1\\ \ln \left(\frac{x^2}{y^2}\right)+\ln \left(\frac{x}{y-x}\right)& =\ln x+c_1\end{array}$$

Applying exponential to each side gives

$$\begin{array}{}\text{(1)}& \left(\frac{x^2}{y^2}\right)\left(\frac{x}{y-x}\right)=c_{2}x\end{array}$$

Singular solution is when $\frac{u(1-u)}{3u-2}=0$. This gives $u=0$ and $u=1$. Hence this implies $y=0$ and $y=x$. Therefore the solutions are

$$\begin{array}{rl}\left(\frac{x^2}{y^2}\right)\left(\frac{x}{y-x}\right)& =c_{2}x\\ y& =0\\ y& =x\end{array}$$

Lets say that we had also initial conditions $y\left(1\right)=-1$, then the above gives

$$\begin{array}{rl}\left(\frac{1}{-1-1}\right)& =c_2\\ -\frac{1}{2}& =c_2\end{array}$$

Therefore the solution (1) becomes

$$\left(\frac{x^2}{y^2}\right)\left(\frac{x}{y-x}\right)=-\frac{1}{2}x$$