1.3.5 Example 5 \(y=x+p^{2}\) (d’Alembert)

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

\begin{align} y & =x+p^{2}\tag {1}\\ & =xf+g\nonumber \end{align}

Hence \(f\left ( p\right ) =1,g\left ( p\right ) =p^{2}\). Since \(f\left ( p\right ) \neq p\) then this is d’Alembert 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 \(f=1,g=p^{2}\) the above simplifies to

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

The singular solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(p-f=0\) or \(p-1=0\). Hence \(p=1\).  Substituting these values of \(p\) in (1) gives singular solution as

\begin{equation} y=x+1 \tag {3}\end{equation}

General solution is found when \(\frac {dp}{dx}\neq 0\,\). Eq (2A) is a first order ode in \(p\). Now we could either solve ode (2) directly as it is for \(p\left ( x\right ) \), or do an inversion and solve for \(x\left ( p\right ) \). Since (2) is separable as is, no need to do an inversion. Solving (2) for \(p\) gives

\[ p=\operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1 \]

Substituting this in (1) gives the general solution

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

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