Stationary path of functional, possible Transversality condition?

Hi everyone,

I have the following functional

**S[y]** **= $\displaystyle \int_{0}^{v}\! \left( {\frac {\rm d}{{\rm d}x}}y \left( x \right) \right) ^{2}+ \left( y \left( x \right) \right) ^{2}\,{\rm d}x$**

Where **y(0) = 1, y(v) = v, v > 0**,

I then showed that the stationary path of the functional is **y = cosh x + B sinh x, 0 ≤ x ≤ v**, by solving the differential equation to obtain **y = A cosh x + B sinh x**, and then using my condition of **y(0) = 1** to prove that **A = 1**.

I am now trying to show that **v = cosh v + B sinh v **and that **B**^{2} − 1 = 2 sinh v + 2B cosh v

For the first part can I just put my condition of **y(v) = v** into my equation to obtain **v = cosh v + B sinh v**?

For the second part I really dont know where to start.... from looking at a similar question i suspect it might be something to do with a Transversality condition, but i'm really not sure how to do this.

Any ideas?

Thanks