Euler- Lagrange, Several functions of 1 variable Question

Question

In this question here, by solving the E-L equations for y and z, you get that $y'' = z$ and $z'' = y$.

Thus $y'''' = y$ and $z'''' = z$

However, this solution is $ Ae^x + Be^{-x} + C\sin x + D\cos x$ and there are 4 variables for only 2 equations.

Which step have I gone wrong at?

Any help would be greatly appreciated.

Thanks!

Answer 1

Everything is as it should be. $y''=z$ and $z''=y$ link the free parameters for the two functions, so there's only one set of $4$ parameters left, and you have $4$ initial conditions for them.