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.

Hint: Let $y'=z$. Then the equation can be written as $z'=a\sqrt{1+z}$. We recognize this as a separable equation that can be solved using standard techniques. 
And now that you have $z$, that is, $y'$, you have another first-order equation to solve.
Added: For this I really prefer Leibniz notation. So we are looking at 
$$\frac{dz}{dx}=a\sqrt{1+z},$$
which can be written as
$$\frac{dx}{\sqrt{1+z}}=a\,dx.$$
Integrate. We get 
$$2\sqrt{1+z}=ax+C.$$
Now replace $z$ by $y'$, solve for $y'$, and continue.
