Setting up and solving for ODE

[hotcommodity]

[SOLVED] Setting up and solving for ODE

Homework Statement

A 5 gallon bucket is full of pure water. Suppose we begin dumping salt into the bucket at a rate of .25 lbs per minute. Also, we open the spigot so that .5 gallons per minute leaves the bucket, and we add pure water to keep the bucket full. If the salt water solution is always well mixed, what is the amount of salt in the bucket after 1 minute?

Homework Equations

If I call the rate of salt flowing in $$\frac{dI}{dt}$$ then $$\frac{dI}{dt}= \frac{.25 lb salt}{min}$$

If I call the rate of solution flowing out $$\frac{dO}{dt}$$ then $$\frac{dO}{dt}= \frac{0.5 gal solution}{min}$$

The concentraion as the amount of salt at any time t divided by the amount of pure water is $$\frac{S(t)}{5gal}$$. Multiplying dO/dt by the concentration gives $$\frac{dO}{dt} = \frac{0.1S}{min}$$.

The Attempt at a Solution

I figure $$\frac{dS}{dt} = \frac{dI}{dt} - \frac{dO}{dt}$$

and thus $$\frac{dS}{dt} = .25 - 0.1(S)$$

I won't type out the rest until I know that I'm on track. Is this correct reasoning?

Any help is appreciated.

 

[Mathdope]

I think you are on the right track, yes.

 

[hotcommodity]

I think you are on the right track, yes.

Ok then. I have:

$$\frac{dS}{dt} = .25 - 0.1(S) = 0.1( 2.5 - S)$$

I separate variables (really only S) and integrate:

$$\int \frac{dS}{2.5 - S} = \int 0.1 dt$$

$$ln|2.5 - S| = 0.1t + C_1$$

$$2.5 - S = \pm e^{0.1t}e^{C_1}$$

$$2.5 - S = C_2e^{0.1t}$$

$$S(t) = 2.5 + C_3e^{0.1t}$$

If there is 0 salt at t = 0, then C3 = 2.5, and I have

$$S(t) = 2.5 + 2.5e^{0.1t}$$

However, this is incorrect.

 

[Mathdope]

If there is 0 salt at t = 0, then C3 = 2.5, and I have

$$S(t) = 2.5 + 2.5e^{0.1t}$$

However, this is incorrect.

C3 would be -2.5 not 2.5

 

[hotcommodity]

But at 1 min, that would give me a negative answer.

 

[Mathdope]

It's possible that the problem lies in the fact that you're derivative measuring rate in is in terms of salt (with no water), while the derivative measuring rate out is in terms of the salt solution (with the water).

If they are adding pure water to keep the bucket full, you may need to recheck what you put for dI/dt.

Edit: No , now that I look again, it does look like both rates are measuring that of the salt only. Hmmm. I'll keep looking.

 

[hotcommodity]

OK thanks, I appreciate the help.

Edit: If it helps at all the answer is 0.238 lbs.

 

[Mathdope]

Try factoring out a -0.1 instead of a 0.1 when you are setting up the differential equation. It shouldn't matter (!) but it will give you a negative exponential instead of a positive. I have to go eat but I'll be back in a half hour or so to see if that helps. Maybe someone else will see the problem too.

Edit: see below. It does matter. That's your problem right there, as "explained" below.

 

[Mathdope]

Ok then. I have:

$$\frac{dS}{dt} = .25 - 0.1(S) = 0.1( 2.5 - S)$$

I separate variables (really only S) and integrate:

$$\int \frac{dS}{2.5 - S} = \int 0.1 dt$$

$$ln|2.5 - S| = 0.1t + C_1$$

It is not the case that d/dS(ln|2.5 - S|)=1/[2.5 - S], so your antiderivative is a bit off (not much). Can you see why?

 

[hotcommodity]

It is not the case that d/dS(ln|2.5 - S|)=1/[2.5 - S], so you're antiderivative is a bit off (not much). Can you see why?

I believe I should have used u-substitution while integrating. If I had done so, I would have:

$$- ln|2.5 - S| = 0.1t + C_1$$

$$-|2.5 - S| = e^{0.1t}e^{C_1}$$

$$-(2.5 - S) = \pm e^{0.1t}e^{C_1}$$

$$S - 2.5 = C_2e^{0.1t}$$

$$S(t) = C_2e^{0.1t} + 2.5$$

When S(t) is 0, C2 must be -2.5.

$$S(t) = 2.5 - 2.5e^{0.1t}$$

This is pretty much what I had before, I'm not sure what I'm doing wrong.

 

[Mathdope]

I believe I should have used u-substitution while integrating. If I had done so, I would have:

$$- ln|2.5 - S| = 0.1t + C_1$$

$$-|2.5 - S| = e^{0.1t}e^{C_1}$$

This is pretty much what I had before, I'm not sure what I'm doing wrong.

Your mistake is going from the first line to second on the left hand side. In other words
$$- ln|2.5 - S|$$ doesn't become $$-|2.5 - S|$$ when exponentiating. Try it once more.

 

[hotcommodity]

Ok I got it, I needed to place the negative on the other side of the equation. So I have:

$$S(t) = 2.5 - 2.5e^{-0.1t}$$

And I get the correct answer. Thank you very much for your help :)

 

[Mathdope]

Aye that's it. You're welcome. You can now mark the thread solved.