Basically the law of Henry will be valid for each inert gas_{i} separately:
(well, well, except the region, what specialists call "Poyinting effect", i.e.:
very high pressures and very complex organic molecules ...)
q_{i} = l_{i} * p_{i} * V_{k}
Q = ∑ q_{i} 
(50) 
The sum encloses all inert gases_{i} and as well all compartments_{k}
Variable  Definition 

q_{i}  amount of diluted inert gas i within a compartment with volume V_{k} 
l_{i}  solubility coefficient of inert gas_{i} valid for this volume V_{k} 
p_{i}  partial pressure of inert gas_{i} 
V_{k}  volume of compartment k 
For the sum of all partial pressures we will have the law of Dalton:
P = ∑ p_{i} = p_{1} + p_{2} + p_{3} + ...  (51) 
Thus we will have for a stationary state in a compartment the following:
P_{t}(0) = ∑p_{t, i}(0) = p_{t, N2}(0) + p_{t, He}(0)  (52a) 
as well for the alveoli:
P_{alv}(0) = ∑p_{alv, i}(0) = p_{alv, N2}(0) + p_{alv, He}(0)  (52b) 
as well for all times t:
P_{tiss}(t) = P_{t}(t) = ∑ p_{tiss, i}(t) = p_{tiss, N2}(t) + p_{tiss, He}(t)  (52c) 
Thus we expand the Haldaneequation (6a):
P_{t}(t) = P_{alv0,N2} + [P_{t0,N2}  P_{alv0,N2}] e^{kN2t} + P_{alv0,He} + [P_{t0,He}  P_{alv0,He}] e^{kHet}  (53) 
We have to check if this expansion will satisfy the basic differential equation (1a). So we expand (1a) with an additional term for a second inert gas:
dP_{t}(t)/dt = k_{1} * [P_{alv,N2}(t)  P_{t,N2}(t)] + k_{2} * [P_{alv,He}(t)  P_{t,He}(t)]  (54) 
We determine the proportionality factors k_{1} and k_{2}, by calculating d P_{t}(t) / dt from (53) :
dP_{t}(t)/dt =  k_{N2} * [P_{t0,N2}  P_{alv0,N2}] e^{kN2t}  k_{He} * [P_{t0,He}  P_{alv0,He}] e^{kHet}  (53a) 
... and with this terms we replace the left hand side of (54):
 k_{N2} *
[P_{t0,N2}  P_{alv0,N2}]
e^{kN2t}
 k_{He} *
[P_{t0,He}  P_{alv0,He}]
e^{kHet} =
k_{1} * [P_{alv,N2}(t)  P_{t,N2}(t)] + k_{2} * [P_{alv,He}(t)  P_{t,He}(t)] = k_{1} * P_{alv,N2}(t)  k_{1} * P_{t,N2}(t) + k_{2} * P_{alv,He}(t)  k_{2} * P_{t,He}(t) 
(55) 
For the p_{t, i}(t) we exploit once again (52c) resp. (53), put this into (55), right hand side, rearrange and recieve:
 k_{N2} *
[P_{t0,N2}  P_{alv0,N2}]
e^{kN2t}
 k_{He} *
[P_{t0,He}  P_{alv0,He}]
e^{kHet} =
k_{1} * P_{alv,N2}(t) + k_{2} * P_{alv,He}(t)  k_{1} * [P_{alv0,N2} + [P_{t0,N2}  P_{alv0,N2}] e^{kN2t}  k_{2} * [P_{alv0,He} + [P_{t0,He}  P_{alv0,He}] e^{kHet}

(55a) 
We multiply the terms of the squared brackets [ ... ] of the right hand side and rearrange according to constant resp. timedependant terms:
 k_{N2} *
[P_{t0,N2}  P_{alv0,N2}]
e^{kN2t}
 k_{He} *
[P_{t0,He}  P_{alv0,He}]
e^{kHet} =
k_{1} * P_{alv,N2}(t) + k_{2} * P_{alv,He}(t)  k_{1} * P_{alv0,N2}  k_{1} * (P_{t0,N2}  P_{alv0,N2})e^{kN2t}  k_{2} * P_{alv0,He}  k_{2} * (P_{t0,He}  P_{alv0,He})e^{kHet} = k_{1} * [P_{alv,N2}(t)  P_{alv0,N2}] + k_{2} * [P_{alv,He}(t)  P_{alv0,He}]  k_{1} * (P_{t0,N2}  P_{alv0,N2})e^{kN2t}  k_{2} * (P_{t0,He}  P_{alv0,He})e^{kHet} 
(55b) 
We take now the identical boundary conditions for the both inert gases which guided us to the solution of (6a):
P_{alv,N2}(t) = P_{alv0,N2} for all times t and analogous
P_{alv,He}(t) = P_{alv0,He} the equation (55a) will be identically satisfied with: k_{1} = k_{N2} as well: k_{2} = k_{He} 
(55c) 
Thus the approach (53) satisfies the central differential equation (1a),
if the second gas is applied as purely additive:
this yields for the stationary part, i.e. the part with no varation in time in the
alveolar gas (52b) and as well for the timedependant part (52c).
NB: the figures are only a first, coarse approximation: the identification of the variables is for getting an overview over equation (21):
t =  τ / ln2 * ln[ (P_{t}(t)  P_{alv0}) / (P_{t0}  P_{alv0}) ] 
The actual deco time t_{d} is the maximal value out of all calculated deco times which have been assessed for the complete group of all the compartments.
For the terms we have the following:
P_{t.tol.}ig = P_{amb} / b + a 
These a & bvalues are constants and could be looked up in a table. Thus the P_{t}(t) can be calculated for every compartment.
a^{*} = a (He + N_{2}) = [( P_{t, He} * a_{He} ) + ( P_{t, N2} * a_{N2})] / ( P_{t, He} + P_{t, N2} )  
b^{*} = b (He + N_{2}) = [( P_{t, He} * b_{He} ) + ( P_{t, N2} * b_{N2})] / ( P_{t, He} + P_{t, N2} )  
λ^{*} = λ (He + N_{2}) = [( P_{t, He} * λ_{He} ) + ( P_{t, N2} * λ_{N2})] / ( P_{t, He} + P_{t, N2} )  (56) 
As well we have for our two inert gases He and N_{2} (52a, 52b, 52c) and thus:
P_{t.tol.}ig = P_{amb} / b^{*} + a^{*} 
If you take now for the calculation of the deco times t_{d} the equation (21) with all of the above
cited extensions for mixed gases,
you generate an equation with two more unknowns,
i.e.: p_{tiss, N2}(t) und p_{tiss, He}(t) for the time t = t_{d}
because of: P_{t}(t) = p_{tiss, N2}(t) + p_{tiss, He}(t),
und this for all compartments_{k}!
As
a^{*}, b^{*} and as well λ^{*} have been assesssed for time t = t_{0}
, but this is only the start of the deco phase!
At first, t_{d} is unknown and thus as well the respective compartment saturations
p_{tiss}(t_{d}) and thus consequently also a^{*}, b^{*}, λ^{*} at
time t = t_{d}.
P_{t}(t + Δt) = p_{tiss, N2}(t + Δt) + p_{tiss, He}(t + Δt), derived from the previous P_{t}(t), and comparing the
tolerated ambient pressure P_{ambient, tolerated} of the desired depth / decostage depth
and the P_{tissue, tolerated} in question:
is after n timesteps
P_{t}(t = t_{d}) = P_{t}(t_{0} + n * Δt) < = P_{tiss, tol}, we have found with t = t_{d} the required decotime.
Further Possibilities:
P_{t}(t) = P_{alv0,N2} + P_{alv0,He} +
[P_{t0,N2}  P_{alv0,N2}]
e^{kN2t}
+
[P_{t0,He}  P_{alv0,He}]
e^{kHet}
= P_{amb} / b^{*} + a^{*} 
Now we ar useing the taylor expansion for the exponentialterm within the P_{t}(t)
with x =  k_{i} * t, and for e^{x} :
e^{x} ≅ 1 + x + (x^{2} / 2) + (x^{3} / 6) + ... + ... 
We stop the taylor expansion after the 3rd or even already after the 2nd term: this is sufficiently accurrate for small x, that is: short deco stops!
The P_{t}(t) in dependancy with the b^{*} and a^{*} means that the time t appears as well in the numerator AND the denominator and thus we get a cubic or biquadratic polynomial for t in its fullest beauty. This one we have to solve as well numerically. But thus we wouldn't gain anything in comparison to method 1!
(*) a couple here means exactly: more than 1,900 profiles from the TECrecreational set.
We took 6 depth profiles from 30  80 m, 5 bottom times from 20  60 min., and helium
fractions from 5  80 % (16 mixes), over all 480 profiles, and the whole then 4 (four!) times ...
This error increases with the decompression obligation, i.e. with increasing divingdepth and bottomtime and the dependancy on the helium fractions f is, roughly speaking like this, only qualitative, sketch:
We are seeing the decotimes t_{D} sketched over the heliumfractions f_{He}:
the deviations are the differences from equation (17) (= method 2) and the meanvalues out of all the 3 softwares
in comparison to the numerical values according method 1.
I.e. as pointed out above: in the beginning we have this overestimation of deco times (error < 0),
in the region of the regular TECREC trimixes, especially at around 35  45 % He the error converges > 0,
and, by increasing f He is growing steadily (error > 0).