Calc 2 differentials

  • Thread starter tnutty
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  • #1
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1
The air in a room with volume 180 m^3 contains 0.20% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 3 m^3/min and the mixed air flows out at the same rate.

(a) Find the percentage of carbon dioxide in the room as a function of time.




attempt :

dy/dx = rate in - rate out

rate in = 0.05 * 3

rate out = 3 * y/180

from that my answer turned out to be :

y(t) = - (e^ (-t/60) * e ^(ln(9) + 1/300 ) - 9 )

any help?
 

Answers and Replies

  • #2
HallsofIvy
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What help are you asking for? If you want to know if that is the correct answer, no, it is not. For one thing, when t= 0, y should be 0.2(180)= 36 but your formula gives y(0)= eln(9)+ 1/300- 9. If you want to know what you did wrong, we will have to see what you did first!
 

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