4.9.1.1.1 Example 1  \(yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}=1\)

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

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=pp^{\prime }\). Hence the ode becomes

\begin{align*} ypp^{\prime }-p^{2} & =1\\ p^{\prime } & =\frac {1+p^{2}}{p}\frac {1}{y}\end{align*}

This is separable.

\begin{align*} p^{\prime }\frac {p}{1+p^{2}} & =\frac {1}{y}\\ \frac {p}{1+p^{2}}dp & =\frac {1}{y}dy\\ \int \frac {p}{1+p^{2}}dp & =\int \frac {1}{y}dy\\ \frac {1}{2}\ln \left ( p-1\right ) +\frac {1}{2}\ln \left ( p+1\right ) & =\ln y+c \end{align*}

Or, assuming \(p-1>0,p+1>0\)

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

Hence

\[ p=\pm \sqrt {1+c_{2}y^{2}}\]

Therefore the solution to the original ode is

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

This is first order ode which is separable. The first one gives

\begin{align*} y^{\prime }\left ( x\right ) & =\sqrt {1+c_{2}y^{2}}\\ \frac {dy}{\sqrt {1+c_{2}y^{2}}} & =dx\\ \int \frac {dy}{\sqrt {1+c_{2}y^{2}}} & =\int dx\\ \frac {1}{\sqrt {c_{2}}}\ln \left ( \sqrt {c_{2}}y+\sqrt {1+c_{2}y^{2}}\right ) & =x+c_{3}\\ \ln \left ( \sqrt {c_{2}}y+\sqrt {1+c_{2}y^{2}}\right ) & =\sqrt {c_{2}}x+\sqrt {c_{2}}c_{3}\end{align*}

Where \(c_{2},c_{3}\) are constants. Similar solution result for the negative ode.