Toronto Math Forum
MAT2442018F => MAT244Tests => Thanksgiving Bonus => Topic started by: Victor Ivrii 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
$$y = xy’ + \sqrt{(y')^2+1}.$$

Here is my solution for Question 7.

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 $y=h(x)$.

here is my solution

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.

$y =xy'+\sqrt{(y')^2+1}$
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 $x$

$y =xy'+\sqrt{(y')^2+1}$
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 $p =\pm\frac{x}{\sqrt{1x^2}}$,
Sub into $y = \frac{x}{p}$, we can get that:
$y = \pm\sqrt{1x^2}$