Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the equation $$ y = xy' + \sqrt{1 + y'^2} $$ with a one-parameter family of solutions being $$ y = cx + \sqrt{1 + c^2} $$ how would I go about showing that a relation $x^2 + y^2 = 1$ defines a singular solution of the equation on the interval $(-1,1)$?

If someone could break down the question into simpler terms, it would be great too!

share|cite|improve this question
Shouldn’t there be a differential equation somewhere? – Haskell Curry Jan 24 '13 at 0:03
Pardon me. I have just added it in. The whole question appears very obfuscated to me, perhaps due to my inexperience, and i did not realise that the earlier part of the question was relevant. – Deyang Jan 24 '13 at 0:08
Note btw that it's only $y = \sqrt{1-x^2}$ (i.e. the upper semicircle) that's a solution, not the lower semicircle $y = -\sqrt{1-x^2}$. There are no real solutions in the region $-1 \le x \le 1$, $y < \sqrt{1-x^2}$. – Robert Israel Jan 24 '13 at 1:15
up vote 3 down vote accepted

Let $y$ denote the positive solution to $x^2+y^2=1$ on the interval $(-1,1)$, so that $y$ is a definite function and is differentiable. Now use implicit differentiation to get $2x+2yy'=0$, and solve to get $y'=-x/y$. Note that $$\sqrt{1+y'^2}=\sqrt{1+x^2/y^2}=\sqrt{(x^2+y^2)/y^2}=\sqrt{1/y^2}=1/y,$$ where we used $x^2+y^2=1$ at one step, and the assumption $y>0$ when the squareroot dropped out.

Now we're done if this simplified squareroot matches up with $$y-xy'=y-x(-x/y)=y+(x^2/y)=(x^2+y^2)/y=1/y,$$ where we again used $x^2+y^2=1$ at one step. Of course in both simplifications we've replaced $y'$ by its implicit derivative $-x/y$ found earlier.

share|cite|improve this answer

You need to show that there is always a point $x_0$ such that your general solution for some constant $c$ is tangent to the singular solution (this is the definition of the singular solution, it is an envelope curve). To prove that such point exists, you need to show that $$ y_g(x_0)=y_s(x_0),\quad y_g'(x_0)=y_s'(x_0), $$ where $y_g$ is the general solution, and $y_s$ is the singular solution. From your formulas you have that $$ y_g'(x_0)=c $$ and $$ y_s'(x_0)=-\frac{x_0}{y(x_0)}. $$ Equating these and putting in $y_g$ you will find that $$ y(x_0)=-\frac{x_0^2}{y(x_0)}+\sqrt{1+\frac{x_0^2}{y^2(x_0)}}, $$ which is, after some simplification and using the fact that $x_0^2+y^2(x_0)=1$, is an identity, which proves that $x^2+y^2=1$ is a singular solution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.