# Welcome back.

## Da warning.

This problem requires us to solve a first-order linear differential equation. This type of problem, and this method of solving it, requires "only" a basic understanding of integration, and would be demonstrated in the first week or so of an undergraduate class in differential equations.

## Da work.

To put this in a framework that we are used to, suppose we add antifreeze at the rate of 1 quart per unit of time. Then all we have to do is find how long it takes to get to our desired concentration and we automatically get our amount.

Let Q = "the number of quarts of antifreeze in the system at any point in time."

Remember our previous discussion? We start with Q = 4 quarts of antifreeze in the system. We want to see how much antifreeze we have to add to end up with Q = 6 quarts of antifreeze in the system.

First, we need to look at the rate of change of Q, which we call Q' or dQ/dt. That rate of change of the amount of antifreeze is determined by how much antifreeze we add minus how much antifreeze we take away. Sound familiar? Anyway, if 1 drop of antifreeze goes in, 1 drop of mixture goes out. That 1 drop of mixture contains Q/12 drops of antifreeze.

We now have our differential equation in dQ/dt. We want an equation in Q. We use a carefully timed integration to get rid of the dQ and the dt.

 dQ/dt = 1 - Q/12 1 drop in minus Q/12 drops out dQ/dt = -(1/12)(Q-12) factor out -(1/12) from RHS [1/(Q-12)] dQ = -(1/12) dt multiply both sides by dt/(Q-12) ln (Q-12) + C1 = -(t/12) + C2 integrate both sides ln (Q-12) = -(t/12) + C2 - C1 move constants to RHS Q-12 = e-[t/12]+C2-C1 raise e to both sides Q-12 = eC2-C1 e-t/12 adding exponents = multiplying Q-12 = C e-t/12 eC2-C1 is just a constant Q = 12 + C e-t/12 add 12 to both sides

Now all we have to do is figure out what the constant C is. When t = 0, Q = 4, so

 4 = 12 + C e0 C = -8 since e0 = 1, just subtract 12 from both sides

So, our final solution to the initial value problems is:

Q = 12 - 8e-t/12

We want to know how much time must elapse (and therefore how much antifreeze we will need to add) before Q = 6. So we substitute that in and solve for t:

 6 = 12 - 8e-t/12 6 = 8e-t/12 subtract 12 from both sides and multiply by -1 6/8 = e-t/12 divide both sides by 8 8/6 = et/12 invert both sides, since e-m is really 1/(em) ln (8/6) = t/12 take the natural log of both sides t = 12 ln (8/6) multiply both sides by 12 t = 12 ln (4/3) simplify the fraction (engineers do not seem to like to do this)

You may recall that t is the amount of time, but also the number of quarts of antifreeze. Mathematicians will be quite happy with this answer. For all you engineers out there, the answer is 3.4521848694213711292706280719259 quarts, more or less.

## Da reality check.

We cannot really check our work, at least not in the way Ray proved that his original answer was correct. We can, however, make educated guesses about whether our answer of 3.45 quarts seems right.

First, let's look at our answer. We could expect the answer to be more than 3, since we already know it takes 3 quarts if we let out the mixture all at once. I would not expect it to require a lot more, though. So 3.45 looks good.

Second, we want to make sure that it works when t = 0. At that point, we have not added any extra antifreeze, and Q = 12 - 8 = 4. So far, so good.

Finally, we can look at what happens if we wait a long time. We would expect that the amount of antifreeze would approach, but never quite equal, 12. As t becomes very large, e-t/12 becomes very small, so Q which equals 12 - 8e-t/12 approaches, but never quite reaches, 12. Bingo.

So, our three reality checks confirmed what we expected. So from that standpoint, the answer looks right.

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