# 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 $\lim_{t\to\infty}x(t)=\frac52$.

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.

• The last part you edited in was how I learned separation of variables. Since $\frac{dy}{dx}$ isn't actually a fraction, the notion of multiplying by $dx$ is hard to justify until you integrate and change the variables. – user161694 Jul 9 '14 at 2:09

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.

• The units are fine. Half a gallon of water leaves the tank. This is a tenth of the volume of the tank, and the tank is well-mixed, so a tenth of the amount of salt in the tank leaves with the water. – Ian Jul 8 '14 at 19:35