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anonymous
 5 years ago
trying to solve by separate equations this differential: dC/dt = rkC  C is a concentration, r is a constant rate and t is time obviously.
anonymous
 5 years ago
trying to solve by separate equations this differential: dC/dt = rkC  C is a concentration, r is a constant rate and t is time obviously.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have dC/ (r  kC) = dt , correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let u = kC , du = k dC, so we have integral 1/u * du/k = integral t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/k int u / du = int t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln u = kt + c1 , c1 is just a constant, not the concentration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kC r = Ae^(kt) by multipling both sides by 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0C = r/k + A/k e^(kt) , notice that it should be A , but since its a constant it doesnt matter

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i made a mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok back to this step 1/k ln u = t + c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln u = k t + k C oh nevermind

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shadow i tried to take derivative of C = r/k + A/k e^(kt)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0C = [r A e^(kt) ] / k This is final general solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you need an initial condition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0separate variables and integrate: dC/dt = r  k·C <=> 1/(rk·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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont use c as your constant of integration, since we already use C for concentration pick a different letter
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