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Victor Ivrii:
Clairaut Equation
is of the form:
\begin{equation}
y=xy'+\psi(y').
\label{eq1}
\end{equation}
 To solve it we plug $p=y'$ and differentiate equation:
\begin{equation}
pdx= pdx + \bigl(x\varphi'(p) +\psi'(p)\bigr)dp \iff dp=0.
\label{eq2}
\end{equation}
Then $p=c$ and
\begin{equation}
y=cx +\psi(c)
\label{eq3}
\end{equation}
gives us a general solution.

(\ref{eq1}) can have a singular solution in the parametric form
\begin{equation}
\left\{\begin{aligned}
&x=-\psi'(p),\\
&y=xp +\psi(p)
\end{aligned}\right.
\label{eq5}
\end{equation}
in the parametric form.


Problem.
Find general and singular solutions to
$$y = xy’ + \sqrt{(y')^2+1}.$$

Chengyin Ye:
Here is my solution for Question 7.

Victor Ivrii:
Joyce, after you found the general solution $y= cx +\sqrt{c^2+1}$, you find a singular one either in the parametric form $x=f(p), y=g(p)$ or, if possible, as in this case, you exclude $c$ and get $y=h(x)$.

Ming Tang:
here is my solution

Victor Ivrii:
After you got a singular solution
$$
\left\{\begin{aligned}
&x = -\frac{p}{\sqrt{p^2+1}},\\
&y= \frac{1}{\sqrt{p^2+1}},
\end{aligned}\right.
$$
you need to express $p$ through $x$ and then plug it to $y$, getting rid of $y$ completely.

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