The curve $y=y(x)$ passing through the point $(\sqrt{3},1)$ and defined by the following property $$\int_{0}^{y}\frac{f(v)dv}{\sqrt{y-v}}=4\sqrt{y}$$ where $f(y)=\sqrt{1+\frac{1}{y^{'2}}}$ is a part of a
$1.$ Straight line.
$2.$ Circle.
$3.$ Parabola.
I have no idea how to solve this problem. Please give me some hint or solution of the problem. Answer is given $1$st option. Thank a lot.
Considered Volterra equation can be written as an equation for convolution $$f(y)*\dfrac1{\sqrt y} = 4\sqrt y,$$ or $$\mathcal L\left\{f(y\right)\}(s)\mathcal L\left\{y^{-1/2}\right\}(s)=4\mathcal L\left\{y^{1/2}\right\}(s),$$ where $\mathcal L(s)$ is Laplace transform.
Using the formula $$\mathcal L\{t^a\}(s) = \dfrac{\Gamma(a+1)}{s^{a+1}}\text{ for }a=-\dfrac12\text{ and }a=\dfrac12$$ and taking in attention that $$\Gamma\left(\dfrac12\right)=\sqrt\pi,\quad \Gamma\left(\dfrac32\right)=\dfrac{\sqrt\pi}2,$$ we have $$\mathcal L\left\{y^{-1/2}\right\}(s)=\sqrt{\pi}s^{-1/2},$$ $$\mathcal L\left\{y^{1/2}\right\}(s)=\dfrac{\sqrt{\pi}}2s^{-3/2}.$$ Therefore, $$\mathcal L\left\{f(y\right)\}(s)=\dfrac2s,$$ $$f(y)=2.$$
So $$\sqrt{1+\dfrac1{y'^2}}=2,$$ $$y'(x)=\pm\dfrac1{\sqrt3},$$ $$y(x)=\pm\dfrac1{\sqrt3}x,$$ and taking in attention the point $(\sqrt3,1),$ $$y(x)=\dfrac1{\sqrt3}x.$$
So the right answer is $1$ - "straight line".