Calculus of Variations - more confusion

[insynC]

I asked a question earlier about Calculus of Variation, but the question I gave didn't really highlight my confusion well. I've come across some other questions that I think reveal my misunderstanding.

Homework Statement

Solve the Euler equation for the following integral:

(integral from x1->x2) ∫[(y')² + √y]dx

Homework Equations

Euler equation:

∂F/∂y - d/dx (∂F/∂y') = 0

The Attempt at a Solution

F = (y')² + √y

So ∂F/∂y = 1/[2√y] and d/dx (∂F/∂y') = 2y'

Thus: y'' = 1/[4√y]

Although it has been some time since I took an ODE course, I think that the equation above is non-trivial to solve. So either I'm mistaken and this is easy to solve or I'm going about the calculus of variations method mistakenly.

With a similar problem, ∫[1+yy']²dx, I was only able to reduce it down to: y''y² + y(y')² = 0, which again I couldn't solve.

I'm not looking for answers, just want to know where I'm applying the method incorrectly, or if in fact I'm missing a far easier way to apply it.

Thanks for any help!

 

[insynC]

Any thoughts?

Is y'' = 1/[4√y] easily solvable? If not how else can the method be applied?

Cheers

 

[Dick]

As far as calculus of variations goes, they look fine. Just because you get a difficult ODE doesn't mean the method is wrong.

 

[gabbagabbahey]

Any thoughts?

Is y'' = 1/[4√y] easily solvable? If not how else can the method be applied?

Cheers

Try solving for y' first by noting that

$$y''(x)=\frac{d}{dx}y'(x)=\left(\frac{d}{dy}\frac{dy}{dx}\right)y'(x)=\frac{dy'}{dy}y'$$

 

[insynC]

Thanks gabbagabbahey, that was a big help.