Modeling salt and water with differential equations 
From Differential Equations (Blanchard, Devaney, and Hall, page 35).
My question is about the model. I let $x$ be the amount of salt and $t$ the minutes passed. Then
$$\frac{dx}{dt}=\frac14-\frac1{10}x$$
Some algebra gives me
$$\frac{dx}{dt}=\frac{5-2x}{20}$$
and I integrate
$$\int\frac{20}{5-2x}dx=\int\,dt$$
to obtain $-10\ln|5-2x|=t+C$. After some algebra, and keeping in mind that $x(0)=0$, I get
$$|5-2x|=5e^{-\frac{t}{10}}$$
I have two sets of solutions: $x=\frac52\left(1-e^{-\frac{t}{10}}\right)$ and $x=\frac52\left(1+e^{-\frac{t}{10}}\right)$.
At this point I'm confused about what to do. Either model provides "sensible" answers, in that the amount of salt I have is positive, as we might expect. So which one do I use?
Solution: (With help from the Ian's answer.) We have an implicit solution $|5-2x|=Ce^{-\frac{t}{10}}$ which implies that $5-2x=Ae^{-\frac{t}{10}}$ where $A=\pm C$. Then we use our initial condition to find that $A$ is restricted to $5$, so we have $5-2x=5e^{-\frac{t}{10}}$, so that our explicit equation for salt is $x=\frac52\left(1-e^{-\frac{t}{10}}\right)$.
To answer questions (a), (b), through (d) requires a calculator and plugging in. To answer (e) we take the limit $\lim_{t\to\infty}x(t)=\frac52$.
 A: So before anything else, your second solution has $x(0)=5$, so that can't be right.
As for what really happened, you got the general implicit solution:
$$-10 \ln |5-2x| = t + C$$
Since $x(0)=0$, the left side starts out at $-10 \ln(5)$, with the positive logarithm. It will not cross over into the other one; if that ever happened, from continuity your division by $5-2x$ would become a division by $0$ which would be invalid. So you'll have the positive logarithm the entire time. Accordingly:
$$-10 \ln(5-2x) = t + C \\
-10 \ln(5) = C \\
-10 \ln(5-2x) = t - 10 \ln(5) \\
10 \ln(5-2x) = 10 \ln(5) - t \\
10 \ln((5-2x)/5) = -t \\
(5-2x)/5 = e^{-t/10}$$
etc.
The cleanest explanation, which also avoids the issue of treating differentials as separate variables, is to write
$$\frac{20}{5-2y} \frac{dy}{ds} = 1$$
and then integrate both sides with respect to $s$:
$$\int_0^t \frac{20}{5-2y} \frac{dy}{ds} ds = \int_0^t ds$$
then change variables on the left side:
$$\int_{x(0)}^{x(t)} \frac{20}{5-2y} dy = \int_0^t ds$$
Then change variables again, and twiddle some minus signs:
$$\int_{5-2x(t)}^{5-2x(0)} \frac{10}{u} du = \int_0^t ds$$
Now the integral on the left side is over a positive range of $u$, since $5-2x(0)=5>0$, so we wind up with a positive logarithm on the left side.
A: Keeping in mind that $x(0) = 0$ you might wish to rethink the statement that $\frac{5}{2}\left( 1+ e^{-t/10} \right)$ hwich has $x(0) = 5$ is "sensible."
As an aside, you also may wish to recheck your units: A pound of salt is not equivalent to a gallon of water.
