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A very large tank initially contains $100$L of pure water. Starting at time t=$0$ a solution with a salt concentration of $0.6$kg/L is added at a rate of $7$L/min The solution is kept thoroughly mixed and is drained from the tank at a rate of $5$L/min. How much salt is in the tank after $60$ minutes?

I have seen similar problem but I don't understand the way in which you solve this. It seems like most of the problems online skip the steps in between or you need a paying account to see how to solve the problem. So I'm hoping someone here can help me.

I've calculated that $\frac{dy}{dt} = 4.2-\frac{5y}{(100+2t)}$.

But don't know where to go from here. I've seen that I am supposed to calculate for a constant C, but I don't understand how you do that. All help is truly appreciated!

katara
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  • you can solve this DE using an integrating factor - see this https://www.mathcentre.ac.uk/resources/uploaded/mathcentre-ode.pdf – David Quinn Sep 21 '21 at 10:18
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    I asked a very similar question here: https://math.stackexchange.com/questions/4226725/differential-equation-mixing-salt-into-water-problem-where-there-is-also-leakag Take a look at the question and answer and see if it is helpful. – Adam Rubinson Sep 21 '21 at 10:18

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You obtained the right differential equation

$ \displaystyle \frac{dy}{dt} + \frac{5y}{(100+2t)} = 4.2$

This is of the form $y' + f(t) y = c$

$f(t) = \frac{5}{100+2t} ~$ and the integrating factor is given by,

$ I = \displaystyle e^{\int f(t) ~ dt} = e^{\frac{5}{2}\ln(50+t)} = (50+t)^{5/2}$

So multiplying ODE by $I$,

$(50+t)^{5/2} dy + \frac{5}{2} (50+t)^{3/2} y ~ dt = 4.2 (50+t)^{5/2} dt$

$ d[(50+t)^{5/2} y] = 4.2 (50+t)^{5/2} ~ dt$

Now integrate both sides and use the initial condition of $y = 0$ at $t = 0$ to find constant.

Math Lover
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  • I don't quite understand how to find the constant when I have two integrals.. – katara Sep 21 '21 at 14:46
  • Integrate both sides and we can add constant to RHS. So it should be $(50+t)^{5/2} y = 4.2 \times \frac{2}{7} (50+t)^{7/2} + C$. Now take $(50+t)^{5/2}$ to RHS. – Math Lover Sep 21 '21 at 15:20