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So for my homework I've gotten an incorrect answer on this problem 3 times in a row. Here's an overview of my work

A large tank holds 250 liters of water with a salt concentration of 7 grams per liter. A brine solution containing 3 grams per liter is added to the tank at a rate of 9 liters per minute. The well-mixed solution is pumped out of the tank at a rate of 5 liters per minute.

How much salt is in the tank after 15 minutes? Enter your answer to the nearest 0.0001 grams.

\begin{align*} S(t)&= \text{concentration of salt as a function of time.}\\ S'&=27-\left(\frac{5S}{250+4t}\right)\\ I&=(250+4t)^5\\ [S(250+4t)^5]'&=27(250+4t)^5\\ S(250+4t)^5&=\frac{9}{8}(250+4t)^6+C\\ S&=\frac{\frac{9}{8}(250+4t)+C}{(250+4t)^5}\\ S(0)&=1750=\frac{\frac{9}{8}(250)+C}{(250)^5}\\ C&=1468.75(250)^5\\ S(15)&=\frac{\frac{9}{8}(250+4(15))+1468.75(250)^5}{(250+4(15)^5}\\ S(15)&=849.7520 \end{align*}

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    @Andrin: $I$ stands for "Integrating factor", not "Identity function" (the identity function is the function that maps every $x$ to itself).$I$use $\mu$ in my answer because that's what I'm used to, and it's harder to confuse with $1$.2011-04-05

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This is a pretty standard "mixing problem." You went wrong in a couple of places:

  • Your set-up for S'(t) has a wrong sign. (This get spontaneously "fixed" later on, which suggests and error in copying somewhere).
  • But more seriously: Your integrating factor is incorrect.

You start pretty well: if we let $S(t)$ be the amount of salt (in grams) in the tank at time $t$ ($t$ measured in minutes). (Note, $S(t)$ is the amount, not the concentration; your formulas clearly view $S$ as the amount, not the concentration; see Pickahu's set-up if you want to use the concentration instead).

In these problems, the amount of salt at any given time is changing by the formula $\frac{dS}{dt} = \binom{\text{rate}}{\text{in}} - \binom{\text{rate}}{\text{out}}.$ And the initial condition $S(0)$ depends on the problem.

The initial condition is simple enough: you are told there are 250 liters of water, with a salt concentration of 7 grams per liter. So $S(0) = \left(250\ \text{liters}\right)\left(7\ \frac{\text{grams}}{\text{liter}}\right) = 1750\ \text{grams of salt.}$

What about the rates in and out? We are adding 9 liters per minute, each liter containing 3 grams of salt. That is, the rate in is: $\text{rate in} = \left(3\frac{\text{grams}}{\text{liter}}\right)\left(9\frac{\text{liters}}{\text{minute}}\right) = 27\frac{\text{grams}}{\text{minute}};$

What is the rate out? We are letting out 5 liters per minute; each liter will have as much salt as the concentration at time $t$. The concentration at time $t$ is given by the amount of salt at time $t$, which is $S(t)$, divided by the amount of liquid at time $t$.

From the moment we start with $250$ liters, each minute you add $9$ liters and you drain $5$ liters, for a net total addition of $4$ liters per minute. So at time $t$, the total amount of liquid in the tank is $250+4t$. So the concentration of salt at time $t$ is $\frac{S(t)}{250+4t}\ \frac{\text{grams}}{\text{liter}}.$

Since we are draining five liters at this concentration, we have that $\text{rate out} = \left(5\ \frac{\text{liters}}{\text{minute}}\right)\left(\frac{S(t)}{250+4t}\ \frac{\text{grams}}{\text{liter}}\right) = \frac{5S(t)}{250+4t}\ \frac{\text{grams}}{\text{minute}}.$

So the differential equation we need to solve is: $\frac{dS}{dt} = 27 - \frac{5S}{250+4t}.$

Writing this in the standard form, we have S' + \frac{5}{250+4t}S = 27. We need an integrating factor. Letting $\mu(t)$ stand for this factor, multiplying through we have \mu(t)S' + \frac{5\mu(t)}{250+4t}S = 27\mu(t) and we want to realize the left hand side as the derivative of a product; that is, we want \mu'(t) = \frac{5\mu(t)}{250+4t}. Separating variables we have \begin{align*} \frac{\mu'(t)}{\mu(t)} &= \frac{5}{250+4t}\\ \int\frac{d\mu}{\mu} &= \int \frac{5\,dt}{250+4t}\\ \ln|\mu| &= \frac{5}{4}\ln|250+4t| + C\\ \mu(t) &= A(250+4t)^{5/4} \end{align*} Picking $A=1$, we can take $\mu(t) = (250+4t)^{5/4}$. (Another error in your computation).

That is, we have: (250+4t)^{5/4}S' + \frac{5(250+4t)^{5/4}}{250+4t}S = 27(250+4t)^{5/4} or (250+4t)^{5/4}S' + 5(250+4t)^{1/4}S = 27(250+4t)^{5/4} which can be written as \Bigl( (250+4t)^{5/4}S\Bigr)' = 27(250+4t)^{5/4}.

You might benefit from writing out the derivations very carefully (as I did above) rather than trying to rely on formulas (I assume that's how you tried to obtain your integrating factor $I$, which was mistakenly computed).

Can you take it from here? Careful with the integral on the right hand side.

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    in the examples worked in class. the difference between output and input was 1. so the problem looked like S'=27-5S/(250-t). This made the integration much simpler.2011-04-05
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I will attempt to give an alternative way to think about this problem, that will make it easier in the future for you. I will not finish the question.

First, note that the accumulation of salt is the amount of salt coming in minus the amount of salt leaving. So, we get that

$ \mathrm{Accumulation}=\mathrm{Salt \, in}-\mathrm{Salt \, out}$

So, which this we try and formulate the equation

$ \mathrm{Volume} \frac{d}{dt} \mathrm{Concentration \, of \, Salt} = \mathrm{flowrate \, in \times concentration \, in} - \mathrm{flowrate \, out \times concentration \, out} $

Then, we can get that $\frac{d(250+4t)S}{dt} = 9 \times 3 - 5 \times S $

The rest should be easy.

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    Thanks for catching that. I was going to do that then it totally slipped my mind as I wrote the solution down.2011-04-05