# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Thanksgiving Bonus => Topic started by: Victor Ivrii on October 05, 2018, 06:00:58 PM

Title: Thanksgiving bonus 6
Post by: Victor Ivrii 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
$$y = xy’ + ( y')^2.$$
Title: Re: Thanksgiving bonus 6
Post by: Jiexuan Wei on October 05, 2018, 10:09:45 PM
Here is my solution. :)
Title: Re: Thanksgiving bonus 6
Post by: YurunyiYang on October 05, 2018, 11:19:03 PM
here is my solution
Title: Re: Thanksgiving bonus 6
Post by: Victor Ivrii on October 06, 2018, 01:40:00 AM
Cathy, the last thing you foub=nd was a singular solution, not a general one!