# Thread: Differential Equation Mixing Problem

1. ## Differential Equation Mixing Problem

I'm having a bit of trouble with some questions, I'm not sure whether it is from deficiancies in D.E or integration and would like some help if possible.

so here's an example.

A tank with 200L of pure water initially, has 4grams of salt per litre run into the tank at 1L per min. A well mixed solution is drained out at 3L per min.

I am to work out the amount of salt in the tank after t mins. Here's what im doing:

V=200-2t--------volume function of time

Dx/Dt=4-(3x/200-2t)------transpose

Dx/dt+(3x/200-2t)=4

integrating factor= e^((integral of)3/200-2t)
=(200-2t)^(-3/2)

Equation then becomes

x(200-2t)^(-3/2)={integral of}4((200-2t)^(-3/2))
Transposing
x =4(200-2t)^(-.5)+c

This equation doesn't seem to be realistic as shouldn't the amount of salt in the container look like a parabola or trig(sin/cos) function?

2. You are absolutely right

$dy/dt$= rate coming in - rate coming out
$dy/dt$ = 1*0.004-3y/200
$dy/dt$ =(-0.015)(y-4/15)
$dy/(y-4/15)$ = $(-0.015)dt$
take integral of both side
$ln|y-4/15|$ = $-0.015t$
$y$ = $e^{-0.015t}+4/15$

3. Originally Posted by rubic
You are absolutely right

$dy/dt$= rate coming in - rate coming out
$dy/dt" alt="dy/dt" /> = 1*0.004-3y/200
$dy/dt$ =(-0.015)(y-4/15)
$dy/(y-4/15)$ = $(-0.015)dt$
take integral of both side
$ln|y-4/15|$ = $-0.015t$
$y$ = $e^{-0.015t}+4/15$

Rate out is=ConcentrationxVolume

C=y/(200-2t) v=3

so shouldn't rate out =3y/(200-2t) as the volume is changing with respect to time?
I should also mention that x, or y as you used is in grams.
thanks

4. so shouldn't rate out =3y/(200-2t) as the volume is changing with respect to time?

oh yes, you are right..
I try to solve again, and i had soln as a straight line
thats what i found
$y = 0.8 - 0.008t$

i am no longer sure about it..