Example 1

\begin{align*} xy^{\prime }-y-2\sqrt {yx} & =0\\ y^{\prime } & =\frac {y}{x}+\frac {2}{x}\sqrt {yx}\end{align*}

For real \(x\)

\begin{align*} \frac {dy}{dx} & =\frac {y}{x}+2\sqrt {\frac {yx}{x^{2}}}\\ & =\frac {y}{x}+2\sqrt {\frac {y}{x}}\end{align*}

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

\begin{align*} x\frac {du}{dx}+u & =u+2\sqrt {u}\\ x\frac {du}{dx} & =2\sqrt {u}\\ \frac {du}{u^{\frac {1}{2}}} & =\frac {2}{x}dx\qquad \sqrt {u}\neq 0 \end{align*}

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

\begin{align*} \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{align*}

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{align*} \sqrt {\frac {y}{x}} & =\ln x+c_{2}\\ y & =0 \end{align*}