Solution to second order second degree equation $ x^2 (d^2y/dx^2)^2=(dy/dx)^2+1$. I am unawae of any method that is there to solve such an equation. IT is totaly different from the usual linear differential equation of degree 2. Any Hint as to how to solve.

Starting from ConMan's answer, we then have $$u_{\pm}=\sinh\left(c\pm\log(x)\right)$$ Using the definition of $\sinh(.)$, we the have $$u_+=\frac 12\left(e^c x-\frac {e^{-c}}x\right)=\frac 12\left(k x-\frac {1}{kx}\right)$$ $$y'_+=u_+\implies y_+=\frac 12\int\left(k x-\frac {1}{kx}\right)\,dx=\frac k4x^2-\frac 1{2k} \log(x)+k'$$ Just do the same for $y_-$
