How do i show that an equation has a singular solution? 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!
 A: 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.
A: 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.
