Hi, this is my first post and I have two past paper questions I'm going over and finding difficulty in the last part.

The 1st question is,

i) If F = y^2 -2xy - (y')^2 , [a,b]=[0,(pi/2)]

y(0)=1

y(pi/2)=0, show that the extremal is

Yo(x)= x -(pi/2)sinx + cosx

now, i done that bit, however, the next part asks for me to find the extremal(s) in which y(pi/2) is not specified.

So, my working has yielded this,

Yo(x) = Acosx +Bsinx + x y(0)=1

y(0)=1 T.F. A=1

so, Yo(x)= x+ cosx +Bsinx

T.C Fy'| x=(pi/2) Fy' = -2y'

t.f. -2y'(pi/2) = 0 t.f. y'(pi/2) = 0

sub this into y'(x)= 1 -sinx + Bcosx

i get 1 -1 + Bcos(pi/2) = 0

the "ones" cancel out and since cos(pi/2)=0, I'm left with 0 = 0 and no value for B.

this is the other question I have had a problems with.

ii) F = (y')^2 + 2y - y^2 [a,b] = [0, pi/2]

y(0)=2

y(pi/2)=0

and this will show that the extremal is

Yo(x) = cosx - sinx + 1

however, the tricky part for me is that I have to write down the transversaility conditions if BOTH y(0) and y(pi/2) are not specifed, and find the extremal in this case.

So, what I have is Yo(x)= Acosx + Bsinx + 1 and

for Fy'| x=0, I get y'(0)=0

Fy'| x= pi/2 I get y'(pi/2) = 0

T.f. y'(0)=y'(pi/2) and t.f.

-Asinx + Bcosx| x=0 = -Asinx + Bcosx| x=pi/2

t.f. B = -A

Or is it that A and B are equal to 0 when plugging these in separetly?

I will appreciate all the help I can get.

Thank you

Jas