How can i solve this second-order differential equation $y^{\prime\prime}=a\sqrt{1+y^{\prime}}$? My problem is to solve this given equation: $y^{\prime\prime}=a\sqrt{1+y^{\prime}}$
My approach was: But i dont know how to handle $a$ and besides this fact, it is a second-order equation. so i thought i'll have to find out, what's $y^{\prime}$ like. But i don't know.
$$y^{\prime}=\int a\sqrt{1+y^{\prime}}\ dx$$ (or $dy$?)
$$y^{\prime}=a\cdot \int \sqrt{1+y^{\prime}}\ dx$$ 
I think i could need help in solving this equation.

Denote $u=y'$, then
$$u'=a\sqrt{1+u}\;\Longrightarrow\; \int\frac{du}{\sqrt{1+u}}=\int adx.$$
This in turn implies
$$2\sqrt{1+u}=ax+C\;\Longrightarrow\; u=\frac{(ax+C)^2}{4}-1.$$
Recalling that $u=y'$ and integrating once more, we find
$$y=\frac{(ax+C)^3}{12a}-x+D,$$
where $C$, $D$ denote two arbitrary constants of integration.
