1.2.3 Example 3 \(y-ay^{\prime }-\sqrt {1+\left ( y^{\prime }\right ) ^{2}}=0\)

Here, we can only isolate \(y\) and use (2A). (note, this can also be solved as dAlembert). Solve

\begin{equation} y-ay^{\prime }-\sqrt {1+\left ( y^{\prime }\right ) ^{2}}=0 \tag {1}\end{equation}

This is problem chapter 7, problem 7. From Boole book, page 137. Let \(y^{\prime }=p\). The above becomes

\begin{align} y-ap-\sqrt {1+p^{2}} & =0\tag {2}\\ y & =f\left ( p\right ) \nonumber \\ & =ap+\sqrt {1+p^{2}}\nonumber \end{align}

Using (2A) gives

\begin{align*} \frac {dx}{dp} & =\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}}\\ & =\frac {a+\frac {p}{\sqrt {1+p^{2}}}}{p}\\ & =\frac {a\sqrt {1+p^{2}}+p}{p\sqrt {1+p^{2}}}\\ & =\frac {a}{p}+\frac {1}{\sqrt {1+p^{2}}}\end{align*}

This is quadrature ode. Solving gives

\[ x=a\ln p+\operatorname {arcsinh}p+c_{1}\]

Hence the parameteric solution is

\begin{align*} y & =ap+\sqrt {1+p^{2}}\\ x & =a\ln p+\operatorname {arcsinh}p+c_{1}\end{align*}

Eliminating \(p\) gives

\[ p=\frac {ya+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\]

Which results in the following implicit solution for \(y\)

\[ x=a\ln \left ( \frac {ay+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\right ) +\operatorname {arcsinh}\left ( \frac {ay+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\right ) +c_{1}\]

Another solution is

\[ p=-\frac {-ya+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\]

Which results in the following implicit solution for \(y\)

\[ x=a\ln \left ( -\frac {-ya+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\right ) -\operatorname {arcsinh}\left ( -\frac {-ya+\sqrt {a^{2}+y^{2}-1}}{a^{2}-1}\right ) +c_{1}\]