Why is $(y-\lambda)y''=1+(y')^2 \tag{1}$ different from usual catenery equation?
How to solve $(y-\lambda)y''=1+(y')^2$?
I obtained (1) by Let $\Gamma$ be a $C^2$ variation with fixed end points parametrized by $s$, i.e. $\Gamma : \prod_{i=1}^n [-\epsilon_i, +\epsilon_i] \to C^{2}([a \ \ b], p, q) $ is a $C^2$ map such that for each $s\in\prod_{i=1}^n[-\epsilon_i, +\epsilon_i] $
\begin{align*}\Gamma_s \in C^{2}([a \ \ b], p, q)= \textit{the space of twice continuously differentiable paths from } p\ \text{to}\ q \\ =\{y\ \vert \ y: [a \ \ b]\to \mathbb{R} \textit{ is a smoothdif curve such that } y(a)=p, \ y(b)=q\}\end{align*}
Then $\Gamma$ can be thought also as $\Gamma :\prod_{i=1}^n [-\epsilon_i, +\epsilon_i] \times [a \ \ b]\to \mathbb{R}$.
Length functional is given as $g:C^{2}([a \ \ b], p, q)\to \mathbb{R}$ $g(y)=\int_a^b\sqrt{1+(\frac{\partial y}{\partial t})^2} dt$
Energy functional is given as $f:C^{2}([a \ \ b], p, q)\to \mathbb{R}$ $f(y)=\int_a^b\Gamma\sqrt{1+(\frac{\partial y}{\partial t})^2} dt$
Given a $C^2$ variation $\Gamma : \prod_{i=1}^n [-\epsilon_i, +\epsilon_i] \to C^{2}([a \ \ b], p, q)$,
define $G: \prod_{i=1}^n[-\epsilon_i, +\epsilon_i] \to \mathbb{R}$ as $ G(s)=g(\Gamma_s)$
define $F: \prod_{i=1}^n[-\epsilon_i, +\epsilon_i] \to\mathbb{R}$ as $F(s)=f(\Gamma_s)$
For a fixed length $c\geq \sqrt{(b-a)^2 + (q-p)^2}$, if the restricted function $F \vert_{G^{-1}\{c\}}$ attains minimum at $s_*$ and if $\vec{\nabla}G\neq0$ there, by Lagrange multiplier theorem, there exists $\lambda\in\mathbb{R}$ such that $\vec{\nabla} F = \lambda \vec{\nabla} G \\ \text{i.e., } \frac{\partial}{\partial s_i} f(\Gamma_s)= \lambda \frac{\partial}{\partial s_i} g (\Gamma_s)$
$\frac {\partial} {\partial s_i} f(\Gamma_s)= \int_a^b\frac{\partial \Gamma}{\partial s}(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{1}{2}}+\Gamma(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-1}{\ 2}} \frac{\partial \Gamma}{\partial t}\frac{\partial^2 \Gamma}{\partial s \partial t} dt$
$\lambda\frac{\partial}{\partial s_i} g(\Gamma_s)= \lambda \int_a^b(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-1}{2}}\frac{\partial \Gamma}{\partial t}\frac{\partial^2 \Gamma}{\partial s \partial t}dt$
\begin{align*} \frac{\partial}{\partial s_i} f(\Gamma_s)-\lambda\frac{\partial}{\partial s_i} g(\Gamma_s) &=\int_a^b\frac{\partial \Gamma}{\partial s_i}(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{1}{2}}+(\Gamma-\lambda)(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-1}{\ 2}} \frac{\partial \Gamma}{\partial t}\frac{\partial^2 \Gamma}{\partial s_i \partial t} dt \\ &=\int_a^b \frac{\partial \Gamma}{\partial s_i}[(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-1}{\ 2}}-\frac{\partial^2 \Gamma}{\partial t^2}(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-3}{\ 2}}(\Gamma - \lambda) ] dt \\ &= \int_a^b \frac{\partial \Gamma}{\partial s_i}(1+(\frac{\partial \Gamma}{\partial t})^2)^{\frac{-3}{\ 2}} [(1+(\frac{\partial \Gamma}{\partial t})^2)-\frac{\partial^2 \Gamma}{\partial t^2}(\Gamma - \lambda) ] dt=0 \end{align*} for any variation if and only if $\begin{align*} (1+(\frac{\partial \Gamma}{\partial t})^2)-\frac{\partial^2 \Gamma}{\partial t^2}(\Gamma - \lambda) =0 \end{align*}$
You can redistribute the factors to $$ \frac{y'y''}{1+y'^2}=\frac{y'}{y−λ}\implies 1+y'^2=C(y−λ)^2 $$ If you have an identity $1+A^2=B^2$ you can parametrize it as $A=\sinh(u)$, $B=\cosh(u)$. Thus set $$ y'=\sinh(u),\\\sqrt{C}(y−λ)=\cosh(u) \\~\\ \implies \sqrt{C}y'=\sinh(u)u'\implies \sqrt{C}\sinh(u)=\sinh(u)u'\implies u'=\sqrt{C}$$ or $$ y(x)=λ+\frac{1}{\sqrt{C}}\cosh(\sqrt{C}x+D). $$