I've tried this problem quite a few times but I can't seem to get it right.
A large tank contains 60 litres of water in which 23 grams of salt is dissolved. Brine containing 15 grams of salt per litre is pumped into the tank at a rate of 8 litres per minute. The well mixed solution is pumped out of the tank at a rate of 2 litres per minute.
(a) Find an expression for the amount of water in the tank after t minutes.
(b) Let x(t) be the amount of salt in the tank after t minutes. Which of the following is a differential equation for x(t)?
EDIT: I totally misread part (a) and thought it was asking for an equation for the amount of salt in the tank. I do still need help finding x(t), since I need it for the next question.
I'm assuming I have to find (b) before I can answer (a). I found the differential equation to be
$$\frac{dx}{dt}=120-\frac{2 x(t)}{60+6t}$$
which was correct. Since this is a linear first order DE, I found the integrating factor
$$m(x)=e^{\int\frac{2}{60+6t} dt}$$
$$m(x)=e^{\frac{1}{3}\int\frac{1}{10+t} dt}$$
$$m(x) = e^{\frac{1}{3} ln(10+t)}$$
$$m(x) = (10+t)^{\frac{1}{3}}$$
Multiplying the DE by this gives
$$\int ((10+t)^{1/3}x(t))'= \int 120*(10+t)^{1/3}$$
$$(10+t)^{1/3}x(t)=120(\frac{3}{4}(10+t)^{4/3}+C)$$
$$\therefore x(t) = \frac{120(\frac{3}{4}(10+t)^{4/3}+C)}{10+t)^{1/3}}$$
Using the inital value $x(0)=23$, since 23 grams of salt is dissolved in the tank initially, I found C to be
$$23 = \frac{120(\frac{3}{4}(10+0)^{4/3}+C)}{10+0)^{1/3}}$$
$$C = \frac{23}{900}$$
Subbing this into the equation and simplifying gives the final answer
$$x(t) = \frac{120(\frac{3}{4}(10+t)^{4/3}+\frac{23}{900})}{(10+t)^{1/3}}$$
$$x(t) = \frac{90(10+t)^{4/3}+\frac{46}{15}}{(10+t)^{1/3}}$$
Which wasn't correct.
EDIT: Also, I should add, I only have one try left on the above question, but the next question is
In Problem #8 above the size of the tank was not given. Now suppose that in Problem #8 the tank has an open top and has a total capacity of 192 litres. How much salt (in grams) will be in the tank at the instant that it begins to overflow?
Since there is 60L in the tank initially, and the contents if increasing at 6 L/min, I found t with the equation
$$60+6t = 192$$
$$t = 22$$
I never actually entered the above equation, so I'm not 100% sure that it's incorrect, but I entered $x(22) = 2880.97$ into this question and it was incorrect.
Best Answer
When you solve for $C$, you should get $$ 23 = \frac{120(\frac{3}{4}(10+0)^{4/3}+C)}{(10+0)^{1/3}} = \frac{120\left(\frac{3}{4} \cdot 10^{4/3}+C\right)}{10^{1/3}} $$ hence, $$ C = \frac{23 \cdot 10^{1/3}}{120} - \frac{3}{4} \cdot 10^{1/3} $$