3.3.7.9 Example 9

\[ xyy^{\prime }=y^{2}+x\sqrt {4x^{2}+y^{2}}\]

Let \(y=ux\) or \(u=\frac {y}{x}\), hence \(y^{\prime }=xu^{\prime }+u\) and the above ode becomes

\begin{align*} x^{2}u\left ( xu^{\prime }+u\right ) & =u^{2}x^{2}+x\sqrt {4x^{2}+u^{2}x^{2}}\\ x^{2}u\left ( xu^{\prime }+u\right ) & =u^{2}x^{2}+x^{2}\sqrt {4+u^{2}}\hspace {0.5in}x>0\\ u\left ( xu^{\prime }+u\right ) & =u^{2}+\sqrt {4+u^{2}}\\ uxu^{\prime }+u^{2} & =u^{2}+\sqrt {4+u^{2}}\\ uxu^{\prime } & =\sqrt {4+u^{2}}\\ u^{\prime } & =\frac {1}{x}\frac {\sqrt {4+u^{2}}}{u}\\ \frac {u}{\sqrt {4+u^{2}}}du & =\frac {1}{x}dx\\ \int \frac {u}{\sqrt {4+u^{2}}}du & =\int \frac {1}{x}dx\\ \sqrt {4+u^{2}} & =\ln x+c_{1}\end{align*}

But \(u=\frac {y}{x}\), hence the above becomes

\begin{align*} \sqrt {4+\frac {y^{2}}{x^{2}}} & =\ln x+c_{1}\\ \sqrt {\frac {4x^{2}+y^{2}}{x^{2}}} & =\ln x+c_{1}\end{align*}

Or for \(x>0\)

\[ \frac {\sqrt {4x^{2}+y^{2}}}{x}=\ln x+c_{1}\]