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trying to solve by separate equations this differential: dC/dt = r-kC ---- C is a concentration, r is a constant rate and t is time obviously.

Mathematics
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you have dC/ (r - kC) = dt , correct?
let u = -kC , du = -k dC, so we have integral 1/u * du/-k = integral t
-1/k int u / du = int t

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Other answers:

-1/k ln u = t + c
ln u = -kt + c1 , c1 is just a constant, not the concentration
so u = Ae^(-kt)
r - kC = Ae^(-kt)
kC -r = -Ae^(-kt) by multipling both sides by -1
C = r/k + A/k e^(-kt) , notice that it should be -A , but since its a constant it doesnt matter
i think i made a mistake
ok back to this step -1/k ln u = t + c
ln u = -k t + -k C oh nevermind
shadow i tried to take derivative of C = r/k + A/k e^(-kt)
C = [r- A e^(-kt) ] / k This is final general solution
now you need an initial condition
separate variables and integrate: dC/dt = r - k·C <=> 1/(r-k·C) dC = dt => ∫1/(r - k·C) dC = ∫dt => -(1/k)·ln(r - k·C) = t + a (a is constant of integration) r - k·C = e^(-k·(t+a)) <=> C = [ r - A·e^(-k·t) ] /k where A= e^(-k·a) use initial condition to evaluate A C(t=0) = C₀ <=> C₀ = [ r - A·e^(-k·0) ] /k => A = r - k·C₀ => C = r/k - (r/k - C₀)·e^(-k·t) (b) the exponential term vanishes for large t. Hence limt→inf C = r/k The concentration in the blood rises from initial value C₀ to r/k
I'm reading through your work at the moment. It makes sense and what I was supposed to see is that the concentration approaches r/k
yeah i figured
dont use c as your constant of integration, since we already use C for concentration pick a different letter
[\int x]
[\x=2\]

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