Question

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.

Answer 1

you have dC/ (r - kC) = dt , correct?

Answer 2

let u = -kC , du = -k dC, so we have integral 1/u * du/-k = integral t

Answer 3

-1/k int u / du = int t

Answer 4

-1/k ln u = t + c

Answer 5

ln u = -kt + c1 , c1 is just a constant, not the concentration

Answer 6

so u = Ae^(-kt)

Answer 7

r - kC = Ae^(-kt)

Answer 8

kC -r = -Ae^(-kt) by multipling both sides by -1

Answer 9

C = r/k + A/k e^(-kt) , notice that it should be -A , but since its a constant it doesnt matter

Answer 10

i think i made a mistake

Answer 11

ok back to this step -1/k ln u = t + c

Answer 12

ln u = -k t + -k C oh nevermind

Answer 13

shadow i tried to take derivative of C = r/k + A/k e^(-kt)

Answer 14

C = [r- A e^(-kt) ] / k This is final general solution

Answer 15

now you need an initial condition

Answer 16

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

Answer 17

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

Answer 18

yeah i figured

Answer 19

dont use c as your constant of integration, since we already use C for concentration
pick a different letter

Answer 20

[\int x]

Answer 21

[\x=2\]