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A tank with a capacity of $500$ gal originally contains $200$ gal of water with $100$ lb of salt in solution. Water containing $1$ lb of salt per gal is entering at a rate of $3\frac{gal}{min}$ and the mixture is allowed to flow out at $2 \frac{gal}{min}$. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow.

https://www.math.ucsd.edu/~c1woods/teaching/tankproblem.pdf

Is this the right way to solve the problem?

What I learn is as below

For mixture problems we have the following differential equation denoted by x as the amount of substance in something and t the time.

$$\frac{dx}{dt}=IN-OUT$$

So, using my book way to solve the above problem! we would have

$$IN=(1)(3) =3$$

So we would have gain of t in each minute

example 3 in, 2 out net=1 (1 minute)

6 in, 4 out (2 minute)

So we would have a net of t

$$OUT=\frac{2x}{200+t}$$

$$\frac{dx}{dt}=3-\frac{2x}{200+t}$$

$$\frac{dx}{dt}+\frac{2x}{200+t}=3$$

Method of integrating factor since this equation is linear!

We would have

$$x(200+t)^2=(200+t)^3+c$$

Can someone explain my way does not work or otherwise!

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Your way works fine. You're missing the initial condition that $x(0) = 100$, from which you can determine $c$.

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  • $\begingroup$ Can you explain to overflowing and the salt concentration to me? $\endgroup$ – Crazy Jun 15 '17 at 13:45
  • $\begingroup$ I figure it out by letting x that I got over the maximum gal! Thanks for posting by the way! $\endgroup$ – Crazy Jun 15 '17 at 13:51
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    $\begingroup$ Right. After you know $c$, then the tank overflows when $t=300$. Concentration is $salt/water$, so in the OUT part, salt is $x$ and water is $200+t$, since each minute the tank gains 1 gallon. $\endgroup$ – B. Goddard Jun 15 '17 at 13:54

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