A question on exponentialize and demoivre

Dear Colleagues,

One more question, please.

I have read in the manuals that 

EXPONENTIALIZE
default: [FALSE] if TRUE will cause all circular and hyperbolic
functions to be converted to exponential form.  (Setting 
DEMOIVRE : TRUE will reverse the effect.) 

I have tried to do this (in the session below), but I failed to
reverse the effect of exponentialize with demoivre for 
(exp(x)+exp(-x))/2 (i.e.to return to cosh(x)) and exp(x) too. 

(C1) display2d : false$
(C2) exponentialize : true$
(C3) e1 : cos(x);
(D3) (%E^(%I*x)+%E^-(%I*x))/2
(C4) e2 : cosh(x);
(D4) (%E^x+%E^-x)/2
(C5) exponentialize : false$
(C6) demoivre : true$
(C7) ev(e1);
(D7) COS(x)
(C8) ev(e2);
(D8) (%E^x+%E^-x)/2
(C9) demoivre(exp(%i*x));
(D9) %I*SIN(x)+COS(x)
(C10) demoivre(exp(x));
(D10) %E^x

What should I do, please, for the transformation of an expression 
containing exponentials to hyperbolic sines and cosines? This is 
important for me in order to have my results in analogous cases 
appear in analogous ways, e.g. in the solutions of the odes 
x''(t)+x(t) = 0 (trigonometric solutions) and x''(t)-x(t) = 0 (hyper-
bolic solutions). I do not like the present difference in notation:

(C11) ode2('diff(x,t,2)-x=0,x,t);
(D11) x = %K1*%E^t+%K2*%E^-t
(C12) ode2('diff(x,t,2)+x=0,x,t);
(D12) x = %K1*SIN(t)+%K2*COS(t)

and I would prefer to use an existing Maxima command (such
as demoivre) avoiding to prepare my own simplification
rules for the second case (that finally leading to hyperbolic
functions). Your advice and help will be greatly appreciated.

Many sincere thanks again for your kind help and support.
I try to get improved, but I need more time (and practice
with Maxima of course!)

Best regards,

Nikos