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!