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!

  • $\begingroup$ too long for here, I'll write an answer shortly. $\endgroup$ – James S. Cook Dec 17 '12 at 0:37

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:

enter image description here

  • $\begingroup$ better go with this answer, I did not check your work ;) $\endgroup$ – James S. Cook Dec 17 '12 at 2:16
  • $\begingroup$ @Artem How did you make this graph? $\endgroup$ – Archisman Panigrahi May 3 '17 at 3:19

First, yes, the solution is not unique. This is a highly nonlinear equation so it's behavior is quite different from the linear DEs we tend to focus upon. Now, the term "general" solution is also unfortunate since clearly the general solutions and singular solutions are distinct. The usual usage of the term "general" would force us to include both the general and the singular solutions into one big solution.

Turning to the other part of your post, pick any point on your singular solution, say $(x_1,y_1)$ then calculate, $$ dy/dx = d/dx (x\ln|x|-x) = \ln|x|+1-1 = \ln|x| $$ Hence the slope to the singular solution at $(x_1,y_1)$ is simply $\ln|x_1|$ hence the particular general solution $y = y_1+\ln|x_1|(x-x_1)$ is tangent to the singular solution at the point in question. This amounts to $c_1=\ln|x_1|$ and $c_2 = y_1-x_1\ln|x_1|$.

Hope this helps.

  • $\begingroup$ I kind of understand what you are saying, but I am having a difficult time envisioning it $\endgroup$ – Cactus BAMF Dec 17 '12 at 1:03

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