Proof that Singular Solution of Clairaut's Equation is the envelope of the family of General Solutions

Question

I want to show that the singular solution is the envelope for the general solutions.

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Proof Outline

1. Both solutions pass from the same point $(a,b)$

2. Both solutions have the same gradient at that point (are tangent to each other)

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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

$$x + f'(y') = 0$$

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.

Answer 1

Comparing ode and its solution

$$y(x,y') = xy' + f(y')$$

$$y (x,C)= xC + f(C)$$

directly establishes $y'=C$, i.e., all associated points $(a,b)$

$$ x=a, y = a x + f( a) = b $$

belong to the envelope.

I suppose that this already completes an adequate proof of the solution.