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.


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.


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




Method of integrating factor since this equation is linear!

We would have


Can someone explain my way does not work or otherwise!


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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