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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?

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    $\begingroup$ (Hint) The numerator looks like the derivative of whats inside the radical $\endgroup$ – Jonathan Davidson Aug 12 '17 at 4:35
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$$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$ $$\frac{y'^2}{(y')^2 + 1} = (x + c)^2 \quad\to\quad y'^2 = \frac{(x + c)^2}{(1-(x + c)^2)}$$ $$y' =\pm \frac{x + c}{\sqrt{1-(x + c)^2}}$$ $$y =\pm \int\frac{x + c}{\sqrt{1-(x + c)^2}}dx=\mp\sqrt{1-(x + c)^2}+C$$ $$(y-C)^2+(x+c)^2=1$$ I suppose that you can take it from here to find the values of $C$ and $c$ according to the specified conditions.

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  • $\begingroup$ I was able to move forward, but one thing scratches my head. how did you get the integral of $\pm \frac{x + c}{\sqrt{1-(x + c)^2}}.$ I was looking at my Table of Integrals, but to no avail. Maybe I will add this to my Table of Integrals? $\endgroup$ – Palautot Ka Aug 21 '17 at 15:58
  • $\begingroup$ This is not a new integral. Just change of variable : $$X=1-(x+c)^2\quad\to\quad dX=-2(x+c)dx \quad\to\quad y=\mp\int \frac{dX}{2\sqrt{X}}$$which is in your table. $\endgroup$ – JJacquelin Aug 29 '17 at 15:22

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