Setting Up a Differential Equation

Let $A$ represent the amount of sugar in the tank. Then:

$\begin{aligned} \frac{dA}{dt} &= (\text{rate of sugar in}) - (\text{rate of sugar out})\\ &= \left(\frac{10 \text{ L}}{\text{min}}\right)\left(\frac{0.2 \text{ kg}}{\text{ L}}\right) - \left(\frac{10 \text{ L}}{\text{min}}\right)\left(\frac{A \text{ kg}}{400\text{ L}}\right)\\ &= 2 - \frac{A}{40}. \ _\square \end{aligned}$

Samir Khan already explained it very well so just for fun, to see what i'd get, i decided to solve this differencial equation, using the method i found on this web site http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx: A'(t)=2-A/40 <=> A'(t)+A 0.05=2 We multiply both sides by the integrating factor, let's name it R(t), where: R'(t)=R(t) 0.05<=>R'(t)/R(t)=0.05, integrate on both sides(knowing that u'/u=(ln u)')to get: ln R(t)=0.05t<=>R(t)=e^(0.05t) So: A'(t) e^(0.05t)+A(t) 0.05 e^(0.05t)=2 e^(0.05t). Now knowing that 0.05 e^(0.05t) is ( e^(0.05t)' and u' v+u v'=(u v)' we can recognise that our last step of the equation can be written as: A'(t) e^(0.05t)+A(t) (e^(0.05t))'=2 e^(0.05t)<=>(A(t) e^(0.05t))'=2 e^(0.05t). Now we do the same process as before when we wanted to find the integrating factor, so we integrate both sides: A(t) e^(0.05t)=80 e^(0.05t)+C<=>A(t)=80+C e^(-0.05t). And since we have the initial condition of A(0)=10, we can solve for C wich equals -70, to get the final equation of: A(t)=80-70*e^(-0.05t), wich is always rising and never reaches 80. PS.:Sorry for all the confusion i still don't quite get how to write on Brilliant.