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var('t')
A = function('A',t)
B = function('B',t)
de1 = diff(A,t) == -k1*A+k2*B
de2 = diff(B,t) == k1*A-k2*B
sol = desolve_system([de1,de2],[A,B],ics=[500,0])
Or conceivably it should correspond to A(500)=0 and leaving B(500) unevaluated, but in that case it should be checked whether that is really what happens, and well-documented.