Author Topic: Thanksgiving bonus 6  (Read 1375 times)

Victor Ivrii

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Thanksgiving bonus 6
« on: October 05, 2018, 06:00:58 PM »
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
$$y = xy’ +  ( y')^2.$$
« Last Edit: October 05, 2018, 06:05:00 PM by Victor Ivrii »

Jiexuan Wei

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Re: Thanksgiving bonus 6
« Reply #1 on: October 05, 2018, 10:09:45 PM »
Here is my solution. :)

YurunyiYang

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Re: Thanksgiving bonus 6
« Reply #2 on: October 05, 2018, 11:19:03 PM »
here is my solution

Victor Ivrii

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Re: Thanksgiving bonus 6
« Reply #3 on: October 06, 2018, 01:40:00 AM »
Cathy, the last thing you foub=nd was a singular solution, not a general one!