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!
Best Answer
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.