Question

A large tank contains 128 gallons of brine in which 10 pounds of salt is dissolved. Brine containing 5 pounds of dissolved salt per gallon runs into the tank at the rate of 10 gallons per minute. The mixture, kept uniform by stirring, runs out of the tank at the rate of 4 gallons per minute. How much salt is in the tank at the end of 13 minutes?

Answer 1

I know its dX/dt=rate in-rate out, but I'm just getting stuck on the actually ODE itself

Answer 2

So right, dx/dt = (rate in)(concentration in) - (rate out)(concentration out). The concentration out is pretty much always the unknown, so just gotta know how to set it up from here. First off, the 10 pounds already dissolved is simply an initial condition, we'll ignore it until the end of the problem. Our first part, rate in and concentration in is simply (5)(10) = 50
The rate out is 4, but the concentration out is the unknown part. The concentration out will be a ratio of the current concentration x over the amount of solution in the tank as a function of time. So the concentration out is then
\[\frac{ x }{ 128-ds/dt }\]. Finding ds/dt is as simple as solving
\[\int\limits_{}^{}(r_{1}-r_{2})dt\], which is integral of 10-4 = 6t, meaning we have the concentration out being
\[\frac{ x }{ 128-6t }\]That is all the information we need to actually set up and do the problem. This is what we currently have:
\[\frac{ dx }{ dt }= 50-\frac{ 4x }{ 128-6t }\]From here it's easy to put this into the form of a linear DE, the form being
\[x' + p(t)x = q(t)\]All thats needed is to move over the 4x term and solve like a linear DE
\[x' + \frac{ 2x }{ 64-3t }= 50\], having just simplified the coefficients in that middle term.

Answer 3

why is it is rate in * concentration in ?
x(t) is the amount of salt at time t.
what are the units of concentration

Answer 4

p(t)=2/3t+64---->>>> P(t)= (3t+64)^2/3?

Answer 5

Because I did all of this. I am getting stuck after distributing the P(t)

Answer 6

The salt is the "concentration." I just didnt actually type units with it. But we have the rate of the brine coming in multiplied with the concentration of salt. Just in generally, I type it out as rate*concentration, could have been someother substance.

As for the p(t) part, in a linear DE, we want:
\[e^{\int\limits_{}^{}p(t)dt}\], which helps us find an integrating factor.

Answer 7

Solving for that integral, we find our integrating factor, which we then use to multiply the whole equation by. Multiplying the whole equation by the integrating factor allows us an easy integration on both sides because the left side of the equation simply turns into us integrating the result of a product rule derivative (which the integrating factor convenient creates), and the right side is usually something pretty basic to integrate anyway.

Answer 8

So for us, we want to do
\[e^{\int\limits_{}^{}\frac{ 2 }{ 64-3t }}dt\]= \[e^{\frac{ -2 }{ 3 }\ln|64-3t|} \implies (64-3t)^{-2/3}\]

Answer 9

what does s(t) stand for?

Answer 10

but wouldn't it be \[\frac{ 2 }{ 64+3t }\] because it is gaining fluid and not losing?

Answer 11

10 gal in- 4 gal out

Answer 12

Yes, +3t, I apologize.

Answer 13

so this is what I get when I do that: \[(3t+64)^{\frac{ 2 }{ 3 }}\]

Answer 14

Distributing that does not seem to give me a linear ode or maybe i'm missing something

Answer 15

It actually does, it just is not maybe as obvious. When you multiply through by the integrating factor, you actually create this kind of form on the left-hand side
\[x't + xt' = q(t)*e^{\int\limits_{}^{}p(t)dt}\], which is the same as \[(xt)' = q(t)*e^{\int\limits_{}^{}p(t)dt}\]You can test it out yourself to check that it actually does a product rule derivative situation, but what happens is now the left side integrates with ease.

Answer 16

\[(3t+64)^{\frac{ 2 }{ 3 }}x'+\frac{ 2x }{(3t+64)\frac{1}{ 3 } }=50(3t+64)\ ^{\frac{ 2 }{ 3 }}\]

Answer 17

This is what I get when I dist.

Answer 18

But I am missing a 3