Integral $\sqrt{1+(dy / dx)^2}= d^2y /dx^2$

Question

A page about the catenary model gives $$ \sqrt{1+ \left(\frac{dy }{ dx} \right)^2}= \frac{d^2 y }{dx^2} $$ and the "integration" leads to $$ \frac{dy}{dx}= \sinh x $$ I donnot understand what happens here, since I find $\cosh x$'s integral is $\sinh x$ but the original equation contains no $\cosh x$. Maybe I misunderstood the symbols.

The original page:

Answer 1

The original equation doesn't need to contain this function explicitly. You can very well verify that $y'=\sinh x$ is a solution as

$$\sqrt{1+y'^2}=\sqrt{1+\sinh^2x}=\cosh x=y''$$ does hold.

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A way to solve this ODE is by setting $y'=z$, which yields

$$z'=\sqrt{1+z^2},$$ or

$$\frac{dz}{\sqrt{1+z^2}}=dx$$ then by integration

$$\text{arsinh } z=x+c$$ and

$$z=\sinh(x+c)=y'.$$

With a second integration, the general solution is

$$y=\cosh(x+c)+c'.$$

Answer 2

It is not hard to verify that $\sinh(x/c)$ is a solution, however if you want to follow the reasoning in your original source, I assume it goes roughly as follows: Start with $$w = H \frac{1}{\sqrt{1+(\frac{dy}{dx})^2}} \frac{d^2y}{dx^2}$$ Now integrate both sides with respect to $x$. Then the left hand side will simply result in $$\int w dx = wx +c_1.$$ The right hand side is a bit more complicated, here you can either substitute $\frac{dy}{dx}$ or see the solution via chain rule. Together with $\frac{d}{dt}\operatorname{arsinh}(t) = \frac{1}{\sqrt{1+t^2}}$ you then get $$\int H \frac{1}{\sqrt{1+(\frac{dy}{dx})^2}} \frac{d^2y}{dx^2} dx = H \operatorname{arsinh}(\frac{dy}{dx})+c_2.$$ In total you end up with $$wx+c_1 = H\operatorname{arsinh}\left(\frac{dy}{dy}\right)+c_2$$ which you can solve for $\frac{dy}{dx}$ in order to get $$\frac{dy}{dx} = \sinh\left(\frac{w}{H}x+\frac{c_1-c_2}{H}\right).$$

Answer 3

Upon the substitution $$u=y'$$ your equation transform to $$u'=\sqrt{1+u^2}$$

Separation of variables implies $$ \frac {du}{\sqrt {1+u^2}}=dt $$ Solving the above we get $$ \sinh ^{-1}u=t+c$$

$$ y'= \sinh(t+c)$$