First Order DE  mixing problem 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.

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