Solution

Here the integrand

L = \frac{1 + y ^{'2}}{y}

has no explicit

x

-dependence. The Beltrami identity therefore gives

L - y^{'} L_{y^{'}} = \frac{1}{y 1 + y ^{'2}} = \frac{1}{R}

for a positive constant

R

. Thus

y^{'2} = \frac{R ^{2} - y ^{2}}{y ^{2}},

whose nonvertical solutions are the semicircles

(x - c)^{2} + y^{2} = R^{2} .

The endpoint

(a, a)

and the prescribed value at

x = b

both lie on

(x - a)^{2} + y^{2} = a^{2},

because

(b - a)^{2} + (2 ab - b^{2}) = a^{2} .

The required positive arc is consequently

y (x) = a^{2} - (x - a)^{2} = 2 a x - x^{2}, a \leq x \leq b .

The condition

b < 2 a

keeps the endpoint above the

x

-axis. The sketch is the descending part of the upper semicircle of radius

a

centred at

(a, 0)

, from

(a, a)

(b, 2 ab - b^{2}) .

Solved by gpt-5.6-sol high.

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