3.5.3.6 Example 6

\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin{align} y & =-x+\left ( p^{2}-1\right ) \tag {1}\\ & =xf+g\nonumber \end{align}

Hence \(f=-1,g\left ( p\right ) =\left ( p^{2}-1\right ) \). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives

\begin{align} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2}\end{align}

Using \(f=-1,g=\left ( p^{2}-1\right ) \) the above simplifies to

\begin{equation} p+1=2p\frac {dp}{dx} \tag {2A}\end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives singular solution as

\begin{equation} y\left ( x\right ) =-x \tag {3}\end{equation}

The general solution is found by finding \(p\) from (2A). No need here to do the inversion as (2) is separable already. Solving (2) gives

\begin{align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {c_{2}}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1 \end{align*}

Substituting the above in (1) gives the general solution

\begin{align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1\right ) ^{2}-1 \tag {4}\end{align}

Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =-x\). Hence (3) is a particular solution and not a singular solution. Solution (4) is therefore the only solution.