Differential equations chemical solutions problem

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SUMMARY

The discussion centers on solving a differential equation related to drug concentration in an organ. The rate of blood flow into and out of the organ is 3 cm3/sec, with a liquid volume of 125 cm3 and an initial drug concentration of 0.2 g/cm3. The correct solution for the concentration of the drug in the organ at time t is given by the formula 0.2(1-e-3t/125), and it takes approximately 28.9 seconds for the concentration to reach 0.1 g/cm3. The key equation used is dx/dt + (3x/125) = 0.6, with proper substitution for initial conditions.

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Homework Statement


newbie here..

Blood carries a drug into an organ at the rate of 3 cm^3/sec and leaves at the same rate. The organ has a liquid volume of 125 cm^3. If the concentration of the drug in the blood entering the organ is .2g/cm^3, what is the concentration of the drug in the organ at time t? After how many seconds will the concentration the drug in the organ reach 0.1g/cm^3?

Answer should be .2(1-e(-3t/125)) and 28.9 seconds

Homework Equations



dx/dt + Fo (x)/Vo = FiCi

The Attempt at a Solution



I used the equation above. Where:

Fo=3cm3/sec
Vo=125cm3
Fi=3
Ci=0

I'm really not sure if I'm doing it right, because i just based on my teacher's previous solutions on other problems

i end up at x = xo e(-3t/125)

but i can't seem to arrive at .2(1-e(-3t/125))
 
Last edited:
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I think it's in your assumption of C_{i}=0 that messes things up. Check to make sure that this is indeed the case.
 
I finally solved it

Instead of assuming Ci = 0, i substituted .2

then I have
<br /> \frac{dx}{dt} + \frac{3x}{125} = (3)(.2)<br />
I used linear differential equation
<br /> v=e^{\int\frac{3}{125} dt} = e^{\frac{3t}{125}}<br />
then
<br /> xe^{\frac{3t}{125}} = \int(.6)(e^{\frac{3t}{125}}) dt + C<br />

then I looked for the value of C which is -.2 and substituted it into the equation

then I finally got x=.2(1-e^{\frac{-3t}{125}})

Thank you very much!
 
Last edited:

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