So I'm trying to do this proof,
The form of Clairauts equation is
$$y(x) = xy' + f(y')$$
You differentiate once to get
$$ y' = y' + xy'' + f'(y')y''$$
You rearrange and get two solutions
The general solution
$$ y = Cx + f(C) $$
The singular solution
$$ x + f'(y') = 0 $$
I want to show that the singular solution is the envelope for the general solutions. For this I need two things:
- Both solutions pass from the same point (a,b)
- Both solutions have the same gradient at that point (are tangent to each other)
From the general solution we get
$$a = (b – f(C))/C $$
Subbing this in to the singular solution we get
$$ b – f(C) + Cf'(y') = 0 $$
Here, I need to show that this equation holds to show that for x=a, the singular solution also passes y=b, which will cover (1) from my requirements.
Also, any help on (2) would be greatly appreciated. I have no idea how to proceed with that.
Best Answer
The envelope of the linear solution family is the curve close to which the lines for $C$ and $C+\Delta C$ intersect (or better the limit of these intersection loci for $ΔC\to 0$). The intersection can be computed as $$ Cx+f(C)=y=(C+ΔC)x+f(C+ΔC) $$ which then implies $$ 0=x+\frac{f(C+ΔC)-f(C)}{ΔC}. $$ In the limit $ΔC\to0$ this is exactly the equation $x(C)+f'(C)=0$. Then $$ y(C)=-f'(C)C+f(C) \implies \frac{dy}{dx}=\frac{y'(C)}{x'(C)} =\frac{-f''(C)C-f'(C)+f'(C)}{-f''(C)}=C $$ as long as $f''(C)\ne 0$.
This can be described geometrically as the line with parameter $C$ having a "focal point" at $(x(C),y(C))$, and the curve of these focal points also having this line as its tangent at this point.