f '(x)=f(2002-x)
f ''(x) = -f '(2002-x)  (1)

Or f '(x)=f(2002-x) ->
f '(2002-x)=f(2002-(2002-x ))
f '(2002-x) = f(x)   (2)

(1) et (2) ->
f''(x) = -f(x)

f ''(x) + f(x) = 0
Equation différentielle du second ordre.
-----
p² = -1
p = +/- i

f(x) = Asin(x) + Bcos(x)
f '(x) = Acos(x) - Bsin(x)
f '(x) = f(2002-x)
Acos(x) - Bsin(x)= A.sin(2002-x) + Bcos(2002-x)

Si x = 0 -> A =  A.sin(2002) +Bcos(2002)
A = B.cos(2002)/(1-sin(2002))
A = -0,402498421831 B
B = -2,48448179114 A

f(x) = Asin(x) + Bcos(x)
f(x) = A.(sin(x) - 2,48448179114.cos(x))
-----
Sauf distraction.