4.9.1.1.10 Example 10 \(yy^{\prime \prime }=2\left ( y^{\prime }\right ) ^{2}+y^{2}\)

\begin{equation} yy^{\prime \prime }=2\left ( y^{\prime }\right ) ^{2}+y^{2} \tag {1}\end{equation}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p\). Hence the ode becomes

\begin{align} y\frac {dp}{dy}p & =2p^{2}+y^{2}\tag {2}\\ yp^{\prime }p & =2p^{2}+y^{2}\nonumber \\ p^{\prime } & =2\frac {p}{y}+\frac {y}{p}\nonumber \end{align}

This is homogeneous class A. Let \(\frac {p}{y}=u\). Hence \(p^{\prime }=u^{\prime }y+u\) and the above becomes

\begin{align*} u^{\prime }y+u & =2u+\frac {1}{u}\\ u^{\prime }y & =u+\frac {1}{u}\\ u^{\prime } & =\left ( \frac {u^{2}+1}{u}\right ) \left ( \frac {1}{y}\right ) \end{align*}

Which is separable as expected. If we do not get separable then we made mistake.  Integrating gives

\begin{align*} \int \frac {u}{u^{2}+1}du & =\int \frac {1}{y}dy\\ \frac {1}{2}\ln \left ( u^{2}+1\right ) & =\ln y+c_{1}\\ \left ( u^{2}+1\right ) ^{\frac {1}{2}} & =c_{1}y\\ u^{2}+1 & =c_{1}y^{2}\end{align*}

But \(\frac {p}{y}=u\), hence

\begin{align*} \frac {p^{2}}{y^{2}}+1 & =c_{1}y^{2}\\ p^{2} & =\left ( c_{1}y^{2}-1\right ) y^{2}\\ p & =\pm y\sqrt {\left ( c_{1}y^{2}-1\right ) }\end{align*}

But \(p=y^{\prime }\), then the above becomes

\[ y^{\prime }=\pm y\sqrt {\left ( c_{1}y^{2}-1\right ) }\]

Hence, looking at one solution

\begin{align*} \int \frac {1}{y\sqrt {\left ( c_{1}y^{2}-1\right ) }}dy & =\int dt\\ -\arctan \left ( \frac {1}{\sqrt {\left ( c_{1}y^{2}-1\right ) }}\right ) & =t+c_{2}\\ \frac {1}{\sqrt {\left ( c_{1}y^{2}-1\right ) }} & =\tan \left ( -t+c_{2}\right ) \\ \sqrt {\left ( c_{1}y^{2}-1\right ) } & =\frac {1}{\tan \left ( -t+c_{2}\right ) }\\ c_{1}y^{2}-1 & =\frac {1}{\tan ^{2}\left ( -t+c_{2}\right ) }\\ y^{2} & =\frac {1}{c_{1}\tan ^{2}\left ( -t+c_{2}\right ) }+\frac {1}{c_{1}}\\ y & =\pm \sqrt {\frac {1}{c_{1}\tan ^{2}\left ( -t+c_{2}\right ) }+\frac {1}{c_{1}}}\end{align*}