Answer
$y = \frac{T}{\rho~g}cosh~(\frac{\rho~g~x}{T})$ is a solution of the differential equation.
Work Step by Step
$y = \frac{T}{\rho~g}cosh~(\frac{\rho~g~x}{T})$
$\frac{dy}{dx} = \frac{T}{\rho~g}sinh~(\frac{\rho~g~x}{T})\cdot \frac{d}{dx}(\frac{\rho~g~x}{T})$
$\frac{dy}{dx} = \frac{T}{\rho~g}~[sinh~(\frac{\rho~g~x}{T})]~\cdot (\frac{\rho~g}{T})$
$\frac{dy}{dx} = sinh~(\frac{\rho~g~x}{T})$
$\frac{d^2y}{dx^2} = cosh~(\frac{\rho~g~x}{T})\cdot \frac{d}{dx}(\frac{\rho~g~x}{T})$
$\frac{d^2y}{dx^2} = (\frac{\rho~g}{T})~cosh~(\frac{\rho~g~x}{T})$
$\frac{d^2y}{dx^2} = \frac{\rho~g}{T}~\sqrt{cosh^2~(\frac{\rho~g~x}{T})}$
$\frac{d^2y}{dx^2} = \frac{\rho~g}{T}~\sqrt{1+sinh^2~(\frac{\rho~g~x}{T})}$
$\frac{d^2y}{dx^2} = \frac{\rho~g}{T}~\sqrt{1+(\frac{dy}{dx})^2}$
Therefore, $y = \frac{T}{\rho~g}cosh~(\frac{\rho~g~x}{T})$ is a solution of the differential equation.