$\displaystyle \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + {(\frac{dy}{dx})^2}}$ ; y(0)=a>0, y'(0)=0

let w = y'

$\displaystyle \frac{dw}{\sqrt{1+w^2}} = \frac{1}{a}dx$

$\displaystyle ln|w + \sqrt{w^2 + a} | = \frac{x}{a} + c$

$\displaystyle ln|y' + \sqrt{y'^2 + a} | = \frac{x}{a} + c$

for the given values c = 0

$\displaystyle y' + \sqrt{y'^2 + a} = e^\frac{x}{a}$

From here I don't know how to proceed. How do you reduce it such that I can get y(x) = ....

Thanks.