### Author Topic: Thanksgiving bonus 6  (Read 2294 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
• Karma: 0
##### Thanksgiving bonus 6
« on: October 05, 2018, 06:00:58 PM »
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
$$y = xy’ + ( y')^2.$$
« Last Edit: October 05, 2018, 06:05:00 PM by Victor Ivrii »

#### Jiexuan Wei

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

#### YurunyiYang

• Newbie
• Posts: 1
• Karma: 2
##### Re: Thanksgiving bonus 6
« Reply #2 on: October 05, 2018, 11:19:03 PM »
here is my solution

#### Victor Ivrii

• Elder Member
• Posts: 2607
• Karma: 0
##### 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!