The final question in a test, for the course, contained this problem. One has perimeter $S : y=y(x)$ on the point $(x,y(x))$ defined by
$$k(x) = \frac{y''(x)}{\left( 1+\left[y'(x)\right]^{2}\right)^{3/2}}\tag{*}$$
and $x\in(-1,1)$, $k(x)=1$, $y(0)=1$ and $y'(0)=0$. There are at least two ways to solve this problem. Now the test instructs to use the way that firstly show the equation in initial value problem:
$$\begin{cases}y'=f(y), &x\in(-1,1) \\ y(0)=1\end{cases}$$
now I am still locked up how to show this form. My first reaction was that I need to integrate that monster two times. Is that the way to simplify the STAR -formula above or how to do that?
Update
I may have actually realized how to solve this problem after writing it down. Hint: use $y(x)=\sinh(x)$ -substitution.
$$y'(x)=\int k(x) dx=\int \cosh^{-2}(x) dx=\frac{1}{2}\left(x+\sinh(x) \cosh(x)\right)_{|y'(x)=\sinh(x)}$$
so
$$\begin{cases}y'=\frac{x}{2}\left(1+\frac{y}{2}\right)^{-1}, &x\in(-1,1) \\ y(0)=1\end{cases}_{|x=Arcsinh(y)}$$
This is an initial value form? Even if it was I find it quite hard form to proceed to the next step where it is supposed to be solved, thinking...