2.26 \(y=xp-e^{p}\) \begin{equation} y=xp-e^{p} \tag {1}\end{equation}
\begin{align} F & =y-xp+e^{p}=0\tag {2}\\ \frac {\partial F}{\partial p} & =-x+e^{p}=0 \tag {3}\end{align}
We first check that \(\frac {\partial F}{\partial y}=1\neq 0\) . Now we apply p-discriminant. Eliminating \(p\) . EQ (3) gives \(e^{p}=x\) or
\[ p=\ln \left ( x\right ) \]
Substituting into the EQ (2) gives
\begin{align*} y-x\ln \left ( x\right ) +e^{\ln \left ( x\right ) } & =0\\ y-x\ln \left ( x\right ) +x & =0\\ y & =x\ln \left ( x\right ) -x \end{align*}
We now have to check if this solution satisfies the ode. We see it does. Now we
have to find the general solution (also called the primitive). This comes out to
be
\begin{align*} \Psi \left ( x,y,c\right ) & =0=y-cx+e^{c}\\ \frac {\partial \Psi }{\partial c} & =0=-x+e^{c}\end{align*}
Eliminating \(c\) . Second equation gives \(c=\ln x\) and Substituting into the first equation above
gives
\begin{align*} 0 & =y-x\ln \left ( x\right ) +e^{\ln x}\\ 0 & =y-x\ln \left ( x\right ) +x\\ y & =x\ln \left ( x\right ) -x \end{align*}
Which is the same as p-discriminant.
Since this is Clairaut, let us use the standard Clairaut ode method to find the singular
solution also and compare. As we know, Clairaut ode has form
\begin{equation} y=xf\left ( p\right ) +g\left ( p\right ) \tag {4}\end{equation}
\(y=xp-e^{p}\) , we see that
\begin{align*} f\left ( p\right ) & =p\\ g\left ( p\right ) & =-e^{p}\end{align*}
The singular solution comes from solving
\begin{equation} x+g^{\prime }\left ( p\right ) =0 \tag {5}\end{equation}
\(p\) and then substituting this back into (4). But
(5) is
\begin{align*} x+\frac {d}{dp}\left ( -e^{p}\right ) & =0\\ x-e^{p} & =0 \end{align*}
Hence
\[ p=\ln \left ( x\right ) \]
Substituting this into (4) gives singular solution
\begin{align*} y & =xp-e^{p}\\ & =x\ln x-e^{\ln x}\\ & =x\ln x-x \end{align*}
Which is same as found using elimination methods.