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Thanksgiving bonus 7

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Victor Ivrii:
Clairaut Equation
is of the form:

y=xy'+\psi(y').
\label{eq1}

To solve it we plug $p=y'$ and differentiate equation:

pdx= pdx + \bigl(x\varphi'(p) +\psi'(p)\bigr)dp \iff dp=0.
\label{eq2}

Then $p=c$ and

y=cx +\psi(c)
\label{eq3}

gives us a general solution.

(\ref{eq1}) can have a singular solution in the parametric form

\left\{\begin{aligned}
&x=-\psi'(p),\\
&y=xp +\psi(p)
\end{aligned}\right.
\label{eq5}

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.