Differential Equations (See photo)

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Differential Equations (See photo)

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Look at this http://www.math.sjsu.edu/~simic/Fall14/Math134/odebook.pdf

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Tank problem? :p Grr I think I've completely forgotten how to do these lol. Gonna read up real quick..
Do you happen to have an answer key? :) I have some steps for you, but I'm not completely confident lol
@zepdrix , I wish I did. That would make it easier... But unfortunately, no I don't :( .
Let A(t) be the `amount of salt in the tank after time t`. The `change in salt amount` is calculated: `the amount coming in` minus `the amount leaving` \[\large\rm A'(t)=A_{in}-A_{out}\] Lemme write it like this I guess...\[\rm A'(t)=\left(\text{salt concentration in}\right)\left(\text{rate of water in}\right)-\left(\text{concentration out}\right)\left(\text{rate of water out}\right)\] Something like that, ya? :d So ummm...
Here is the rate at which salt is coming into the tank,\[\large\rm A_{in}=\left(\frac{0.2kg}{1L}\right)\left(1\frac{L}{\min}\right)\] The first set of brackets shows the concentration of salt to total liquid, while the second set of brackets shows the rate at which it is flowing in. Simplifying gives us:\[\large\rm A_{in}=0.2\frac{kg}{\min}\]
The amount of salt leaving the tank is a little trickier. It starts out as pure water, so we start amount with zero being the amount of salt flowing out. But as time progresses, the concentration builds in the tank, and we get a variable concentration flowing out. What is the concentration that is flowing out? It's going to be the A(t), amount of salt, divided by the total liquid in the tank.\[\large\rm A_{out}=\left(\frac{A(t)kg}{10L}\right)\left(1\frac{L}{\min}\right)\]So again, this will simplify down nicely for us,\[\large\rm A_{out}=\frac{A(t)}{10}\frac{kg}{\min}\]
@Zepdrix, just so you know, I have zero knowledge, at the moment, of ODE, whatsoever. I am trying to see the answers to the problems first, so I know how I should work through these kinds of problems.
Ah ok :) Well I'm pretty sure we're on the right track here. Is this too complicated so far?
I'll take whatever best explanation you give me. :) Just waiting for you to finish, to review it all, actually.
Thanks for your time!
So setting up our equation:\[\large\rm A'=A_{in}-A_{out}\]\[\large\rm A'=0.2\frac{kg}{\min}-\frac{A}{10}\frac{kg}{\min}\]Let's get rid of the units for now, so it's easier to read.\[\large\rm A'=\frac{1}{5}-\frac{1}{10}A\]
If you have zero knowledge of ODE's, then this next couple of steps will probably be a little confusing, but that's ok.
We could find the general solution A(t) by any number of ways.. let's just do the old fashioned integrating factor though.
\[\large\rm A'+\frac{1}{10}A=\frac{1}{5}\]We want to multiply the left side of the equation by something that turns it into product rule.
When we have a general form like this: \(\large\rm y'+p(x)y=q(x)\) (as we do) Then we use an integrating factor, which we denote by mu usually. It has this form:\[\large\rm \mu=e^{\int\limits p(x)dx}\]
So for our equation, the 1/10 is our p(t), nothing more than that.\[\large\rm \mu=e^{\int\limits \frac{1}{10}dt}=e^{\frac{1}{10}t}\]We'll multiply through by our integrating factor,\[\large\rm e^{\frac{1}{10}t}A'+\frac{1}{10}e^{\frac{1}{10}t}A=\frac{1}{5}e^{\frac{1}{10}t}\]Left side is now the result of some product rule, we can reverse it!\[\large\rm \left(A e^{\frac{1}{10}t}\right)'=\frac{1}{5}e^{\frac{1}{10}t}\]
We'll integrate both sides with respect to t,\[\large\rm \int\limits\left(A e^{\frac{1}{10}t}\right)'dt=\int\limits\frac{1}{5}e^{\frac{1}{10}t}dt\]Which gives us:\[\large\rm A e^{\frac{1}{10}t}=2e^{\frac{1}{10}t}+c\]We get a constant of integration which is really important in these types of problems.
Solving for A, dividing both sides by the exponential gives us:\[\large\rm A(t)=2+c e^{-\frac{1}{10}t}\]
We actually have some initial data that we can plug in to solve for c. Recall that the Amount of salt at time t=0 should be zero. There was no salt in the tank to begin with! A(0)=0. So we'll plug this in to solve for c,\[\large\rm A(0)=2+c e^{0}=0\]Which leads to,\[\large\rm 2+c(1)=0\]\[\large\rm c=-2\]Giving us our equation!!\[\Large\rm A(t)=2-2e^{-\frac{1}{10}t}\]
So that answers part A. Part A is the large bulk of the work for the problem. Use can use that equation to answer the other parts.
At time \(\large\rm t=10\ln2\) we have:\[\Large\rm A(t)=2-2e^{-\frac{1}{10}(10\ln2)}\]Do you understand how to simplify this? :o
Woops*\[\Large\rm A(10\ln2)=2-2e^{-\frac{1}{10}(10\ln2)}\]
You only need an understanding of Algebra to do part B and maybe some calc for part C Shouldn't be too bad :) Give it a shot broski!
and i get the same relationship so you have \[\large A'+\frac{1}{10}A=\frac{1}{5}\] \[\large 10A'= 2 - A\] \[\large \int_{0}^{A(t)} \frac{10 \ dA} {2 - A} = \int_{0}^{t} dt\]
@zepdrix , Thanks for staying and working it out. I intended to study your solution a few days ago, but I'm just getting to your solution now. It's been a hectic week. I appreciate your quick response!

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