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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.

neelkanth
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  • I don't understand what $f(y)=\sqrt{1+\frac{1}{y^{'2}}}$ is there for ? In particular the presence of a derivative $y'$... – Jean Marie Aug 29 '16 at 08:51
  • its derivative square..... – neelkanth Aug 29 '16 at 09:21
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    All right, the square of a derivative. But here is not the problem. The problem is that if you write $f(y)$, $y$ is a variable, so what do you mean if you express it by $f(y)=\sqrt{1+\frac{1}{y^{'2}}}$ ? Do you understand what I mean ? And what is the connection with your integral ? – Jean Marie Aug 29 '16 at 09:32
  • Indeed, because of the derivative, the integral on the LHS makes no sense when one injects the expression of $f$ in it (and I find highly annoying that you chose to ignore @JeanMarie's point) . What is the source of this problem? – Did Aug 30 '16 at 12:19
  • It appears, from the accepted answer below, that $f$ is an operator, taking in a differentiable function $y$ and returning a function defined by $x\mapsto \sqrt{1+\frac{1}{y'(x)^2}}$. Can you confirm whether this is the case, for the sake of future readers of this question? – Joey Zou Aug 31 '16 at 01:18

1 Answers1

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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".

Yuri Negometyanov
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