Differential Equation involving Mixture Problem

Question

I have encountered many mixing problem like this but I just can't answer this one. Here is the problem,

A mass of inert material containing $15 \text{ lb}$ of salt in its pores is agitated with $10 \text{ gal}$ of water initially fresh. The salt dissolves at a rate which varies jointly as the number of pounds of undissolved salt and the difference between the concentration of the solution and that of a saturated solution ($3 \text{ lb/gal}$). If $9 \text{ lb}$ are dissolved in $10 \text{ min}$, when will $90\%$ be dissolved?

Here's my solution:
Let $P$ be the amount of dissolved salt, $15-P$ be the undissolved. The concentration of the solution $= \frac P {10} \text{ lb/gal}$. Then, $$\frac{\mathrm dP}{\mathrm dt} = (15-P)\left(3-\frac P {10}\right)\text;$$ $$\frac{\mathrm dP}{\mathrm dt} = (15-P)(30-P)/(10)\text.$$ Since this is separable, I used partial fraction then integrate both side and got an answer of around $11 \text{ min}$, which is wrong according to the textbook. The answer key provided an answer of $30.5 \text{ min}$. I don't really know whats wrong in my equation above.

Answer 1

$\frac{dP}{dt} = k (15-P) (3- \frac{P}{10})$ where $k$ is a constant multiplier.

$dt = \frac{10}{k(15-P)(30 - P)} dP$

$ \displaystyle \int dt = 10 = \frac{1}{k}\int_0^9\frac{10}{k(15-P)(30 - P)} dP = \frac{1}{k} \frac{2}{3} \ln \frac{7}{4}$ (time taken is $10$ mins).

So, $k = \frac{1}{15} \ln \frac{7}{4} \approx 0.0373$

Now for $90\%$ of salt = $13.5$ lb. For it to dissolve, time taken will be

$ \displaystyle \int dt = t = \frac{1}{k}\int_0^{13.5}\frac{10}{k(15-P)(30 - P)} dP = \frac{1.1365}{k}$

So, $t = \frac{1.1365}{0.0373} \approx 30.5$ minutes.

Answer 2

The setup is not quite right. The wording of the problem is a little poor, but ultimately it means that the rate of change of the amount of salt is proportional to the remaining salt and the difference between the concentrations. As such, our equations should read

$$\frac{dP}{dt}=k(15-P)\left(3-\frac{P}{10}\right)$$

Comparing to what you had, we have a proportionality constant and no $dP/dt$ term on the right. The general solution (using separation of variables) is

$$P=30-\frac{15}{1+Ce^{-kt/2}}$$

(can you do this?) There are two conditions we need to satisfy here: namely

$$P(0)=0,\;\;\;\;P(10)=9$$

(Note that we are using $t$ here measured in minutes). The first condition gives us the constant $C$ as

$$0=P(0)=30-\frac{15}{1+C}\implies C=-1/2\implies P=30\left(1-\frac{1}{2-e^{-kt/2}}\right)$$

The second condition gives us $k$, which is

$$9=P(10)=30\left(1-\frac{1}{2-e^{-5k}}\right)\implies k=-\frac{1}{5}\ln\left(\frac{4}{7}\right)$$

Our final equation is then

$$P=30\left(1-\frac{1}{2-\left(4/7\right)^{t/10}}\right)$$

The question asks at what time $t$ is $P$ at 90% of the total amount of salt, so for what $t$ is $P(t)=13.5$. We have

$$13.5=30\left(1-\frac{1}{2-(4/7)^{t/10}}\right)\implies t=\frac{10\ln(11/2)}{\ln(7/4)}\approx 30.46$$

Hope this helps!