4.7.1.1 Example 1

$$\begin{array}{rl}x{y}^{\prime }-y-2\sqrt{yx}& =0\\ y^{\prime }& =\frac{y}{x}+\frac{2}{x}\sqrt{yx}\end{array}$$

For real $x$

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{y}{x}+2\sqrt{\frac{yx}{x^2}}\\ & =\frac{y}{x}+2\sqrt{\frac{y}{x}}\end{array}$$

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

$$\begin{array}{rl}x\frac{du}{dx}+u& =u+2\sqrt{u}\\ x\frac{du}{dx}& =2\sqrt{u}\\ \frac{du}{u^{\frac{1}{2}}}& =\frac{2}{x}dx\sqrt{u}\ne 0\end{array}$$

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

$$\begin{array}{rl}\int u^{\frac{-1}{2}}du& =\int \frac{2}{x}dx\\ 2u^{\frac{1}{2}}& =2\ln x+c_1\\ u^{\frac{1}{2}}& =\ln x+c_2\end{array}$$

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

$$\sqrt{\frac{y}{x}}=\ln x+c_2$$

The singular solution is $u=0$. Which implies $y=0$. Hence the solutions are

$$\begin{array}{rl}\sqrt{\frac{y}{x}}& =\ln x+c_2\\ y& =0\end{array}$$