I think this is going to come up on my exam so I'm worried as I can't do it...

a) If [latex] F = y^2 - 2xy - y'2, y(0) = 1, y(\pi/2) = 0 [/latex] show that the extremal is [latex]y_0(x) = x - \frac{\pi}{2} sin x + cos x. [/latex].

b) Find the extremal in which [latex] y(\pi/2) [/latex] is not specified.

I've can do part a fine, and part b is:

Transversality condition: [latex]Fy' [/latex](at x =b) [latex]= 0. [/latex]

Hence [latex]-2y_0' (\pi/2) = 0 [/latex]

[latex] y_0(x) = cos x + B sin x + x [/latex]

[latex] y_0'(x) = -sin x + B cos x + 1 [/latex]

[latex] 0 = -1 + 1 [/latex]

Hence there are infinitely many extremals.

But for a similar question:

a) If [latex] F = y'^2 + 2y -y^2, y(0) = 2, y(\pi/2) = 0 [/latex] show that the extremal is [latex] y_0(x) = cos x - sin x + 1. [/latex].

b) Write down the transversality condition if both [latex]y(0)[/latex] and [latex]y(\pi/2)[/latex] are not specified.

For part b I don't understand how to do this without any boundary conditions at all...could someone please explain how?

Thanks in advance!