Cannot solve an EL equation for a trivial textbook example Background. Retired engineer learning physics/need to learn calculus of variations as a tool.
Textbook: ML Boas Mathematical Methods for the Physical Sciences. Ed 3 Ch 9 Prob 2.6
Problem: Find $y(x)$ that MAKES 
$$
\int (\frac{dy}{dx})^2 + \sqrt{y(x)} dx
$$
stationary.
I adopt the Euler_Lagrange route and thus need to solve what looks like a simple DE:
$$
\frac{d^2y}{dx^2}-\frac{1}{4\sqrt{y}}=0. 
$$
The ELE functional is given by MLB as 
$$
y'(x)^2+y^{\frac{1}{2}}.
$$
I can't do this by hand (always my first attempt). Mathematica 10 gives a massive, messy solution yet MLB claims that 
$$
x+a = \frac{4}{3}(y^\frac{1}{2} -2b)(b+y^\frac{1}{2})^\frac{1}{2}
$$
'stationary-ides!!!) the ELE integral.
I apologise for this question to a community to which it is probably trivial but I know that mathematics is my toolkit for my study of physics and I must understand where I am going wrong. Thank you in advance for your help, David Mackay.
 A: ${y}''-\frac{1}{4}{{y}^{-\tfrac{1}{2}}}=0\Rightarrow {y}''.{y}'-\frac{1}{4}{{y}^{-\tfrac{1}{2}}}.{y}'=0\Rightarrow \frac{1}{2}\frac{d}{dx}\left( {{({y}')}^{2}} \right)-\frac{1}{2}\frac{d}{dx}\left( {{y}^{\tfrac{1}{2}}} \right)=0$  
Integrating we get
${{\left( {{y}'} \right)}^{2}}-{{y}^{\tfrac{1}{2}}}=b\Rightarrow {y}'={{\left( b+{{y}^{\tfrac{1}{2}}} \right)}^{\tfrac{1}{2}}}$  
The square root taken here may need some thought as the negative root may be appropriate..Taking the positive root and proceeding this is a seperable first order DE and using the substitution $u=b+{{y}^{\tfrac{1}{2}}}$ we get  
$2\int{\frac{(u-b)}{{{u}^{\tfrac{1}{2}}}}}du=\int{dx}=x+a$  
doing the integral then simplifying gives your MLB claim.
A: Since your functional $F = y'^{2} + \sqrt{y}$ doesn't depend explicitly on $x$, there is a first integral of the form
$$\begin{align}
y' \frac{\partial F}{\partial y'} - F &= y' \cdot 2y' - (y'^{2} + \sqrt{y}) \\
&= y'^{2} - \sqrt{y} \\
&= C \\
\implies y'^{2} &= C + \sqrt{y} \\
\implies y' &= \pm \sqrt{C + \sqrt{y}}
\end{align}$$
Hence, separating and integrating gives us
$$\begin{align}
\int \frac{dy}{\sqrt{C + \sqrt{y}}} &= \pm \int dx \\
&= \pm (x + K) \\
\end{align}$$
To solve the LHS, we'll use a change of variable to simplify. Let 
$$\begin{align}
y = u^{2} \implies dy &= 2u du\\
\end{align}$$
Hence
$$\begin{align}
\int \frac{dy}{\sqrt{C + \sqrt{y}}} &= \int \frac{2u du}{\sqrt{C + u}}
\end{align}$$
Using another change of variable, let
$$v = C + u \implies dv = du$$
Hence
$$\begin{align}
\int \frac{2u du}{\sqrt{C + u}} &= 2 \int \frac{(v - C) dv}{\sqrt{v}} \\
&= 2 \bigg[ \frac{2v^{\frac{3}{2}}}{3} - 2Cv^{\frac{1}{2}} \bigg ]
\end{align}$$
Unravelling, we find
$$\begin{align}
2 \bigg[ \frac{2v^{\frac{3}{2}}}{3} - 2Cv^{\frac{1}{2}} \bigg ] &= 2 \bigg[ \frac{2(C + u)^{\frac{3}{2}}}{3} - 2C(C + u)^{\frac{1}{2}} \bigg ] \\
&= 2 \bigg[ \frac{2(C + \sqrt{y})^{\frac{3}{2}}}{3} - 2C(C + \sqrt{y})^{\frac{1}{2}} \bigg ] \\
&= \pm (x + K)
\end{align}$$
