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.
The governing species mass balance on the tank is:
$$ I(t)Q_{in}-c(t)Q_{out}=\frac{d[c(t)V]}{dt}$$ where $Q_{in}$ and $Q_{out}$ are the flow rates in and out of the tank, respectively, which are both equal to 4L/min. Since they're equal so we'll just call them $Q$. $V$ is the tank volume of 1120L and is constant because the inflow and outflow are equal. Thus $V$ will come out of the derivative and we get:
$$ IQ-cQ=V\frac{dc}{dt}. $$ Now we can solve for the desired unknown, $I$:
$$ I(t)=\frac{V}{Q}\frac{dc}{dt}+c. $$ Since $c$ is given, we're done. You can stop here or plug in all the values: take the derivative of $c$ and plug in $\frac{V}{Q}=\frac{1120L}{4L/min}=280min$ to get:
$$ I(t)=\frac{0.35}{80}e^{-t/800}. $$
Hope this helps,
Paul Safier
Best Answer
Substitute $r=20$ into your ode
$$ \frac{\mathrm{d}Q}{\mathrm{d}t} = 0.75r - \frac{Q(10+r)}{2}$$
and rearrange it in the form below,
$$\frac{\mathrm{d}Q}{ 1-Q} =15 \mathrm{d}t$$
Integrate with the initial value $Q(0) =0$
$$ -\ln (1-Q) = 15t$$
and get the solution
$$Q(t)= 1- e^{-15t}$$
The concentration of 0.5 kilogram per liter corresponds to $Q=1$. Then, the time it takes to reach within 5% of $Q=1$ satisfies
$$ 5\% = e^{-15t}$$
Solve to get
$$t= \frac{\ln20}{15}=0.2 \>\text{minute}$$
or, 12 seconds.