Deriving a high ordered Euler-Lagrange equation. I've been able to derive the Euler-Lagrange equation for 
$$\int_a^b F(x,y,y')dx$$ relatively easily by using the total derivative and integration by parts. However, I was unable to apply the same methods whilst trying to derive the Euler-Lagrange for $$\int_a^b F(x,y,y',y'')dx$$ A proof would be great, but a proper name for this type of equation would also be appreciated (without a term I've found scouring the internet for a Euler-Lagrange of this form to be quite difficult).
 A: Start by setting up a variation $y_\epsilon (x) = y(x) + \epsilon h(x)$ where $\epsilon$ is some small real number and $h(x)$ is zero on the boundaries. You now have a functional of the form
$$\int_a^b f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx.$$
Differentiate with respect to $\epsilon$ and use the Leibniz rule.
$$\frac{d}{d \epsilon} \int_a^b f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx = \int_a^b \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx .$$
Note that
$$ \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \epsilon} + \frac{\partial f}{\partial y_\epsilon}\frac{\partial y_\epsilon}{\partial \epsilon} + \frac{\partial f}{\partial y'_\epsilon}\frac{\partial y'_\epsilon}{\partial \epsilon} + \frac{\partial f}{\partial y''_\epsilon}\frac{\partial y''_\epsilon}{\partial \epsilon}$$
It follows that
$$ \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon) =  \frac{\partial f}{\partial y_\epsilon}h(x) + \frac{\partial f}{\partial y'_\epsilon} h'(x) + \frac{\partial f}{\partial y''_\epsilon} h''(x)$$
Your functional now looks like
$$\int_a^b   \left(     \frac{\partial f}{\partial y_\epsilon}h(x) + \frac{\partial f}{\partial y'_\epsilon} h'(x) + \frac{\partial f}{\partial y''_\epsilon} h''(x)     \right)     dx.$$
Since you are looking for a stationary point, set it equal to zero and now evaluate it at $\epsilon = 0$.
$$0 = \int_a^b   \left(     \frac{\partial f}{\partial y}h(x) + \frac{\partial f}{\partial y'} h'(x) + \frac{\partial f}{\partial y''} h''(x)     \right)     dx.$$
Use integration by parts on the second and third terms. I will only perform it on the third term since you state you've derived the simpler case before. This term must be integrated by parts twice.
$$ \int_a^b  \frac{\partial f}{\partial y''} h''(x)        dx.$$
Choose $u=  \frac{\partial f}{\partial y''}$ and $v'= h''(x)  $ so $u' = \frac{d}{dx} \left( \frac{\partial f}{\partial y''} \right)$ and $v = h'(x)$.
The resulting integration by parts is
$$ \frac{\partial f}{\partial y''} h'(x) \bigg|_a^b - \int_a^b \frac{d}{dx} \left( \frac{\partial f}{\partial y''}\right) h'(x) dx$$
The first term is zero since $h(x)$ is zero on the boundaries. The second term must be integrated by parts again but now with
$$ u = \frac{d}{dx}\left( \frac{\partial f}{\partial y''} \right)$$
$$u' = \frac{d^2}{dx^2}\left( \frac{\partial f}{\partial y''} \right)$$
$$v = h(x)$$
$$v' = h'(x)$$ 
I will leave the substitutions to you. The functional now reads:
$$ 0= \int^a_b \left[ \frac{\partial f}{\partial y} - \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) + \frac{d^2}{dx^2} \left( \frac{\partial f}{\partial y''} \right)  \right] h(x) dx $$
This should work for any function $h(x)$ so your Euler-Lagrange equation is
$$ 0=  \frac{\partial f}{\partial y} - \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) + \frac{d^2}{dx^2} \left( \frac{\partial f}{\partial y''} \right) $$
