Finding the Equation of Circle, Given is its Differential Equation I was studying for some quizzes when I stumbled upon  this question.
It goes like this:

Find the equation of the circle whose differential equation is 
  $y'' = (1 + (y')^2)^{\frac{3}{2}}$ and which passes through the points 
  (0, 0) and (1,1).

My work:
The differential equation $y'' = (1 + (y')^2)^{\frac{3}{2}}$ doesn't have a dependent 
variable $y,$ so we let
$$\frac{dy}{dx} = p$$
$$\frac{d^2 y}{ dx^2} = \frac{dp}{dx}$$
Now we had all we need, we can now solve for the general solution of 
$y'' = (1 + (y')^2)^{\frac{3}{2}}$
$$y'' = (1 + (y')^2)^{\frac{3}{2}}$$
$$ \left(\frac{dp}{dx}\right) = (1 + p^2)^\frac{3}{2}$$
$$dp = (1 + p^2)^\frac{3}{2} dx$$
$$\frac{dp}{(1 + p^2)^\frac{3}{2}} = dx$$
$$\int \frac{dp}{(1 + p^2)^\frac{3}{2}} = \int dx$$
Remembering that $\int \frac{du}{(u^2 + a^2)^\frac{3}{2}} = \frac{u}{a^2 \sqrt{u^2 + a^2}} + c$
$$\frac{p}{\sqrt{p^2 + 1}} = x + c$$
$$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$
$$\int \frac{y'}{\sqrt{(y')^2 + 1}} = \int x + c$$
At this point, I'm stuck, because is it possible to do this?
$$\int \frac{y'}{\sqrt{(y')^2 + 1}} = \int x + c$$
How will I go forward, and ultimately, getting the equation of the circle?

$$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$
$$\frac{y'^2}{(y')^2 + 1} = (x + c)^2 \quad\to\quad
y'^2 = \frac{(x + c)^2}{(1-(x + c)^2)}$$
$$y' =\pm \frac{x + c}{\sqrt{1-(x + c)^2}}$$
$$y =\pm \int\frac{x + c}{\sqrt{1-(x + c)^2}}dx=\mp\sqrt{1-(x + c)^2}+C$$
$$(y-C)^2+(x+c)^2=1$$
I suppose that you can take it from here to find the values of $C$ and $c$ according to the specified conditions.
