5.2.1.1 Example 1 \(y^{\prime }y^{\prime \prime \prime }+\left ( y^{\prime }\right ) ^{2}=2\left ( y^{\prime \prime }\right ) ^{2}\)

Let \(u=y^{\prime }\) then

\begin{align*} y^{\prime \prime } & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \\ & =\frac {du}{dx}\\ & =\frac {du}{dy}\frac {dy}{dx}\\ & =u\frac {du}{dy}\end{align*}

And

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

Hence the original ode becomes

\begin{align*} y^{\prime }y^{\prime \prime \prime }+\left ( y^{\prime }\right ) ^{2} & =2\left ( y^{\prime \prime }\right ) ^{2}\\ u\left ( u\left ( \frac {du}{dy}\right ) ^{2}+u^{2}\frac {d^{2}u}{dy^{2}}\right ) +u^{2} & =2\left ( u\frac {du}{dy}\right ) ^{2}\\ u^{2}\left ( \frac {du}{dy}\right ) ^{2}+u^{3}\frac {d^{2}u}{dy^{2}}+u^{2} & =2u^{2}\left ( \frac {du}{dy}\right ) ^{2}\\ \left ( \frac {du}{dy}\right ) ^{2}+u\frac {d^{2}u}{dy^{2}}+1 & =2\left ( \frac {du}{dy}\right ) ^{2}\\ u\frac {d^{2}u}{dy^{2}} & =\left ( \frac {du}{dy}\right ) ^{2}-1 \end{align*}

This is second order ode in \(u\left ( y\right ) \) with missing \(y\).  Let \(\frac {du}{dy}=s\) then \(\frac {d^{2}u}{dy^{2}}=\frac {d}{dy}\left ( \frac {du}{dy}\right ) =\frac {ds}{dy}=\frac {ds}{du}\frac {du}{dy}=s\frac {ds}{du}\). The above becomes

\begin{align*} us\frac {ds}{du} & =s^{2}-1\\ \frac {ds}{du}\frac {s}{s^{2}-1} & =\frac {1}{u}\end{align*}

Which is separable. The solution is

\begin{equation} s=\pm \sqrt {1+c_{1}^{2}u^{2}}\tag {1}\end{equation}

Taking the first solution then

\[ \frac {du}{dy}=\sqrt {1+c_{1}^{2}u^{2}}\]

The solution is

\[ u=\frac {1}{c_{1}}\sinh \left ( c_{1}y+c_{1}c_{2}\right ) \]

But \(u=y^{\prime }\) hence

\[ y^{\prime }=\frac {1}{c_{1}}\sinh \left ( c_{1}y+c_{1}c_{2}\right ) \]

Solving gives

\[ y=\frac {1}{c_{1}}\left ( \operatorname {arccosh}\left ( -\tanh \left ( x+c_{1}c_{3}\right ) \right ) -c_{1}c_{2}\right ) \]

We need to do the same for the other solution in (1)