# $k(x) = \frac{y''(x)}{\left( 1+\left[y'(x)\right]^{2}\right)^{3/2}}$ to initial value problem form?

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...

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...my instructor stated that the problem can be solved with $u(x)=y'$ or $u(y)=y'$. One leads to a solution, the other leads to a mess. Now we have at least 3 different ways, interesting. – hhh Feb 23 '12 at 20:22

The answer boils down to something rather easy! Since you're given that $k=1$, and the expression you show is the curvature of $y(x)$, the problem is solved by any circle with unit radius! This is to say, for any $a$ and $b$, the solution is

$$\tag{1}(x-a)^2+(y-b)^2=1$$

You can also show that the equation you give, with $k(x)=1$ is indeed the equation to $(1)$ by differentiating it twice and elimininating $a$ and $b$ from the equation. Here's how to show it. We differentiate implicitly first.

\eqalign{ & {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = 1 \cr & 2\left( {x - a} \right) + 2\left( {y - b} \right)y' = 0 \cr & \left( {x - a} \right) + \left( {y - b} \right)y' = 0 \cr & \left( {x - a} \right) = - \left( {y - b} \right)y' \cr}

We now substitute back

\eqalign{ & {\left( {y - b} \right)^2}{\left( {y'} \right)^2} + {\left( {y - b} \right)^2} = 1 \cr & {\left( {y - b} \right)^2} = \frac{1}{{1 + y{'^2}}} \cr}

We differentiate $(1)$ once more

\eqalign{ & 1 = - y{'^2} - \left( {y - b} \right)y'' \cr & 1 + y{'^2} = - \left( {y - b} \right)y'' \cr}

We substitute one more time

\eqalign{ & \frac{{{{\left( {1 + y{'^2}} \right)}^2}}}{{y'{'^2}}} = \frac{1}{{1 + y{'^2}}} \cr & 1 = \frac{{y'{'^2}}}{{{{\left( {1 + y{'^2}} \right)}^3}}} \cr & 1 = \frac{{y''}}{{{{\left( {1 + y{'^2}} \right)}^{3/2}}}} \cr}

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You were right in thinking that somehow $\sinh$ enters the game, but what you proposed in your update is not a "substitution"; it is the conjecture that $y(x)=\sinh x$ is the actual solution of the problem.

Looking at the given second order equation one is lead to introduce a new unknown function $x\mapsto u(x)$ via $$y'(x):=\sinh u(x)\ \qquad\bigl(\Rightarrow y''(x)=\cosh u(x)\cdot u'(x)\bigr)\ .$$ In this way the given equation is transformed into a first order differential equation for $u(x)$, namely $$u'(x)=\cosh^2 u(x)\cdot k(x)=\cosh^2 u(x)$$ with the initial condition $u(0)=0$. The latter equation can be separated and solved explicitly, whereupon you have to compute $$y(x):=1+\int_0^x \sinh u(t)\ dt\ .$$ Note that the original second order equation has a certain geometrical interpretation which makes the solution more or less obvious.

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