TITLE: Clairaut's Equation Singular and General Solutions
QUESTION [5 upvotes]: I want to know how one how one would prove that the singular solutions to Clairaut's equation are tangent to the General solutions. so I have here:
$$y(x) = xy' - e ^{y'}$$ Differentiating
$$y' = y' +xy'' - y''e^{y'}$$
$$0 =y''(x-e^{y'})$$
Therefore for the general solution, I have $y'' = 0 \implies y' = c_1 \implies y_g(x) = c_1x + c_2$
Okay thats all well and good. As for the singular solution. I'm still not quite sure how the singular solution differs from the general solution. All I know is it must envelope the family of general solutions as well as be tangent to them (For reasons unknown to me If someone could explain it I would be eternally grateful). The singular solution is found by:
let us create some parameter $$y' = p \implies 0 = x-e^p \implies ln|x| = p$$
Plug this back into the original DE to get the singular solution of y:
$$y_s(x) = xln|x| - x$$
I want to show that one is tangent to the other at some point (By the way a kind of side note, If the two equations exist at the same point, doesn't that mean that the solutions are NOT unique at that point?)
$$y_g(x) = y_s(x) = c_1x + c_2 = xln|x| - x$$
Now what? I solve for some point (x,y)? Then how would I prove that they are tangent? I don't quite understand this part, assuming this is all correct. Thank you anyone for looking at this!

REPLY [5 votes]: First note that your general solution is not entirely correct (you can check it by plugging it into your equation). The correct solution is given by
$$
y_g=Cx-e^C
$$
and depends on only one arbitrary constant.
Your singular solution is correct and given by
$$
y_s=x\ln x-x.
$$
How to show that actually there is always a point at which the general and singular solutions are tangent? Consider the condition that two functions are tangent at a point $x_0$:
$$
y_g(x_0)=y_s(x_0),\quad y'_g(x_0)=y'_s(x_0).\tag{1}
$$
(They should have the same $y$ coordinates and the same slope, and this is exactly what is written there.)
From you solutions and using the second equality in $(1)$ you find that
$$
C=\ln x_0.
$$
Putting this value into the equality $y_g(x_0)=y_s(x_0)$ you will find an identity which proves that your singular solution is tangent to the general solutions.
Here is the picture: