Author Topic: Thanksgiving bonus 7  (Read 1997 times)

Victor Ivrii

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Thanksgiving bonus 7
« on: October 05, 2018, 06:03:59 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
$$y = xy’ + \sqrt{(y')^2+1}.$$

Chengyin Ye

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Re: Thanksgiving bonus 7
« Reply #1 on: October 05, 2018, 11:09:52 PM »
Here is my solution for Question 7.

Victor Ivrii

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Re: Thanksgiving bonus 7
« Reply #2 on: October 06, 2018, 01:45:34 AM »
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

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Re: Thanksgiving bonus 7
« Reply #3 on: October 06, 2018, 02:11:41 PM »
here is my solution

Victor Ivrii

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Re: Thanksgiving bonus 7
« Reply #4 on: October 06, 2018, 05:08:45 PM »
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.

Pengyun Li

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Re: Thanksgiving bonus 7
« Reply #5 on: October 06, 2018, 08:18:45 PM »
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases}
  x = - \frac{p}{\sqrt{p^2+1}} \\
  y = - \frac{1}{p}{x}
 \end{cases}$
No it is not!!! I wrote solution in the parametric form already. Now the question is to exclude $p$ , expressing $y$ via $x$   
« Last Edit: October 07, 2018, 12:36:22 AM by Victor Ivrii »

Pengyun Li

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Re: Thanksgiving bonus 7
« Reply #6 on: October 07, 2018, 01:21:47 PM »
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases}
  x = - \frac{p}{\sqrt{p^2+1}} \\
  y = \frac{1}{\sqrt{p^2+1}}
 \end{cases}$

Since $x = -\frac{p}{\sqrt{p^2+1}}$, we can derive that $p =\pm\frac{x}{\sqrt{1-x^2}}$,

Sub into $y = -\frac{x}{p}$, we can get that:

$y = \pm\sqrt{1-x^2}$