How to solve for the function $f(x)$ such that the area underneath is always equal to the arclength?

Question

First, I set up my equation. The area is obviously $\int f(x) \space dx$. The arclength is $\int \sqrt{1+(\frac{dy}{dx})^2} \space dx$. Thus, I have the equation $\int f(x) \space dx = \int \sqrt{1+(\frac{dy}{dx})^2} \space dx$. Now I differentiatie both sides to get $f(x) = \sqrt{1+(\frac{dy}{dx})^2} $. I will now make some substitutions to make it look simpler like this. $y = \sqrt{1 +y'^2}$. This becomes $y^2 = 1+y'^2$. Since we have a function that is similar to its own derivative, I will guess the solution looks like $y = Ae^{kx}$. $y' = Ake^{kx}$. Pugging these values into our equation, I get $(Ae^{kx})^2 = (Ake^{kx})^2 + 1$ or $A^2e^{2kx} = A^2k^2e^{2kx} + 1$. I am now stuck here, please help.

Answer 1

So you got to $$ y=\sqrt{1+y'^2}\implies 1=y^2-y'^2 $$ This is the equation for a hyperbola, so use hyperbolic coordinates $y=\cosh(u(t))$, $y'=\sinh(u(t))$. The derivative of the first equation combined with the second then implies for $y'\ne0$ that $u'(t)=1$, so $y(t)=\cosh(t+C)$.

The other case $y'(t)=\sinh(u(t))=0$ gives $y=1$, which is also a solution.

Answer 2

The equation you get is going to be $$y' = \sqrt{1+y^2}$$ which is nonlinear, but separable. Integrating, you get $$x = \int \frac{dy}{\sqrt{1+y^2}}$$ and you can solve this with the substitution $y = \sinh (x + C)$.