Recently, I have asked a question about variational analysis (First moments of uniform distribution on a curve from (0,0) to (1,1) in two-space). Such the question can be addressed in some cases by solving the following integral function:
$f(x)=\frac{\lambda\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_1}{\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_2}$, where $f(x)\in\mathbf{C}^1[0,1]$, and $\lambda$, $c_1$ and $c_2$ are constant numbers.
How to solve the above nonlinear equation?
$f(x).\int_{0}^{x} \sqrt{1+{f(t)}^2} dt =c+\lambda =a$
Say, $f(x)=y$
So, $\frac{d (\int_{0}^{x} \sqrt{1+{y(t)}^2} dt)}{dx} =\frac{d\frac{a}{y}}{dx}$
From Newton- Leibniz's formula
$\sqrt{1+y^2}=a\frac{d}{dx}(\frac{1}{y})$
Taking $z=(\frac{1}{y})^2$ , we get
$\sqrt{1+z}=a\frac{dz}{dx}$
or, $\frac{dz}{\sqrt{1+z}}= \frac{dx}{a}$
Or, $y=\frac{1}{\sqrt{(\frac{x}{a}+q)^2-1}}$ , ($q>1$ for $a>1$, $q<1$ for $a<0$)
Putting the result $y=\frac{1}{\sqrt{(\frac{x}{a}+q)^2-1}}$ we can easily see that the integral equation is satisfied.
