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