Point to curve problem: How to solve the Euler's equation?

Question

I'm trying to solve the following problem: Find the extremal curve for $ J(y) = \int_0^{x_1} \frac{ ( 1 + (y')^2)^{\frac{1}{2}}}{y} dx = \int_0^{x_1} F(y,y') dx $ with $y(0) = 0$ and the point $(x_1, y(x_1))$ on the line $ \psi(x) = x -5$. Now, what I know about this problem is that this is the "point to curve minimization problem" and an extremal curve $y$ has to solve the Euler's equation $F_y - \frac{d}{dx} F_{y'} = 0$ plus the transversality condition which, in this cases, is simply $ y'(x_1) = - \frac{1}{\psi' (x_1)} $. The problem is how to solve the Euler's equation: $$ \frac{ \sqrt{ 1 + (y')^2}}{y^2} + \frac{d}{dx} \left( \frac{y'}{y \sqrt{1 + (y')^2}} \right) = 0. $$ I can't find a way to solve this differential equation.

Answer 1

EDIT1:

If $x$ is not explicitly involved in $F$ we can integrate directly the Euler-Lagrange equation itself using Beltrami shortcut

$$ \frac{ \sqrt{ 1 + (y')^2}}{y^2} - \frac{y'}{y}\cdot \frac{y'}{\sqrt{1 + (y')^2}} = C_1 \tag 1 $$

$$ \frac{1}{y{\sqrt{1 + (y')^2}} } =1/C_1 = {C} \text { or } y \sqrt{1+y^{'2} }=C $$

$$\pm \frac{y\; dy }{{\sqrt{C^2 - y^{2}}}} =dx \tag 2 $$

Integrating,

$$ \pm \sqrt{C^2-y^2}= x - h;\, (x-h)^2+y^2 = C^2 $$

where $h$ is a single arbitrary constant, resulting in solution of ODE as all semi-circles centered on x-axis, not fully suitable before incorporating transversality condition $y=x-5$.

No need to go to the second derivative ODE if no transversality condition.We can introduce another degree of freedom by differentiation to obtain all circles in the $(x,y)$ plane to get

$$ \frac{y''}{\left(1 + y^{'2}\right)^{3/2}} = \frac{1}{C}$$

With an extra constraint condition as a requirement to being normal to the straight line: $\; y= x-5 .$

A variable circle centered on

$$ (x_c,y_c)=( \lambda, \lambda-5)$$

with variable subset constant arbitrary radius $C$ ensures orthogonal transversality.. like in the graph: