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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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    Arturo, thanks for cleaning that up. In the future, how do I format in that style?2011-04-05
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    @Andrin: what is $I$?2011-04-05
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    @Andrin: learn $\LaTeX$2011-04-05
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    Your formula for $S'$ is incorrect: every minute, you add 9 liters and you drain 5 liters, so there is a net *gain* of 4 liters per minute. So at time $t$, the amount of liquid in the tank is 250+4t gallons, not 250-4t.2011-04-05
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    picakhu: I believe it stands for Identity function. It is e of the integral of the coefficient of Y [A in this case]. Then you multiply both sides of the equation by I so you get. IY'+ I'Y = CI which simplifies to (IY)'=CI. In this case e^5ln(250+4t) goes to (250+4t)^52011-04-05
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    @Andrin: I think I know that is it. But I think you should be more clear in your steps. Try and show you understand what you are doing rather than you know how to get the answer.2011-04-05
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    @picakhu: I suspect it's the "integrating factor".2011-04-05
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    @Andrin: If you right click on any formula and click on "Show Source", it will show you the LaTeX code; in-line equations need to be enclosed in dollar signs, `$...$`; and displayed formulas in double dollar signs, `$$...$$`. Also, you were missing some parentheses which made some of your expressions ambiguous, and you still have an unbalanced parentheses in $S(15)$.2011-04-05
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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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    I think for this problem, it is more meaningful to write the equation accumulation=in-out. And then explain the terms. That way it is not a mechanical calculation problem, but a concept problem.2011-04-05
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    @picakhu: Good point; edited to include that derivation.2011-04-05
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    thank you. in the examples we went over in class, t did not have a constant. that definitely did me in. I have a DE test tomorrow, so I appreciate your help. The TA's were stumped.2011-04-05
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    @Andrin: I don't understand what you mean by "$t$ did not have a constant." $t$ is the variable, it stands for time.2011-04-05
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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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    @picakhu: The OP is considering $S$ the *amount* of salt (though he calls it the "concentration"); you might want to specify explicitly that you are using $S$ for the concentration (in grams/liter), or use a different letter...2011-04-05
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    @Arturo: the OP specifically said that S is concentration.2011-04-05
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    @picakhu: Yes, you'll notice I mentioned that, so it's not like I did not notice. And then, if you see his formula, you'll see that even though he **calls** it the "concentration", he is really using it as the **amount**, not the concentration. So either he misspoke when he called the "concentration", or else he set everything up completely wrongheadedly. Either way, I thought it would be good to point that out *somehow*, rather than simply go ahead and use $S$ to represent something other than what the OP *actually* used it for without a word to explain the radical difference in equation.2011-04-05
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    @picakhu: Also, since the volume is not fixed at $250$ liters, can you really set it up like this? The amount of salt at time $t$ is not $250S$, it's $(250+4t)S$. So shouldn't the way the amount of salt is changing be $\frac{d}{dt}((250+4t)S) = 9\times 3 - 5\times S$?2011-04-05
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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