Let's do one, the concrete search example. The number of people in a village of size 10,000 people, who are exposed to a certain rumor is governed by the logistic equation with birth rate 3. In letters, I used the letter r, right, small r. Initially, only 10 people hear the rumor. Then, how many days are needed for one half of the whole villages to hear the rumor? Here we take the units of time to be days. So, let small p (t) be the number of people hearing the rumor after t days. Then in this other problem, we have the both rate of 3, r = 3. Initial people who already hear the news, that is a p not, that is equal 10. What is the carrying capacity of this situation? Carrying capacity is a maximum population which is allowed. That should be 10,000 because the village has only 10,000 people. That's the carrying capacity. In symbol, that was r / a. So, r / a is equal to 10 to the 4, 10,000. So what does that mean? That means another constant we need for the logistic equation a, which is equal to 3 x 10 to the -4. So we are ready to find the equation for the p (t). Just plugging all those information, r = 3, p note = 10, and a = 3 times 10 to the -4, into the equation number 5, p(t) = rp note over ap note + (r- ap note) e to the -rt. r = 3, p0 is equal to 10, and so on. Then you will get this equation, p(t) = 3 x 10 over 3 x 10 to the -3 + (3- 3 times 10 to the -3) times e exponent of -3t. Simplify this expression, then you will get 10 to the 4 over 1 + (10 to the 3- 1) times e to the -3t, okay? That is the p(t), the number of people who hear the rumor after t days. What is the problem? How many days are needed for one half of the whole villagers to hear the rumor? How much is the one half of the villager? The total number of villagers are 10,000, so one half of it is a 5,000. That means 5 times 10 to the 3. So here we go, p of t will be 5 times 10 to the 3 after t days. The only thing unknown is the t in this equation. So plug in that, t is unknown. This is simple algebra of solving t from this equation. It gives t is equal to log of 10,000- 1 over 3. In other words, t is equal to log 999/ 3, which is approximately 2.3. In other words, after 2.3 days the half of the whole villages will hear the rumor. Let's consider one another interesting example, the so-called mixing problem. Consider a large tank holding 100 liters of brine solution in which 10 kilograms of salt is dissolved. A brine solution with salt concentration 0.2 kilograms per liter flows into the tank at the rate of 5 liters per minute, and then the well stirred solution flows out of the tank at the rate 3 liters per minute. We'd like to determine the mass of the salt inside of the tank at any time t. Do you understand the problem? For this word problem, where this were the parabola? Let m(t) be the mas of the salt in the tank at time t. Then what's the rate of change of the mass of the salt inside of the tank? In other words, m prime (t), it must be equals to input rate- output rate. And this same m must satisfy the initial condition. Initially, inside of the tank, the problem said we have 10 kilograms of salt, so m(0) = 10. How much is the input rate of salt? The problem said the brine solution is coming into the tank, With the 5 liters per minute, and that brine solution has a salt concentration 0.2 kilograms per liter. That means for every minute 5 times 0.2, in other words, 1 kg of salt is coming in for every minute. That's the input rate of salt into the tank. On the other hand, the problem said well stirred solution inside of the tank is flows out in speed rate 3 liters per minute. Then we need the salt concentration in the tank after t minutes. After t minutes, how many salt do we have inside of a tank? We have m kilograms of salt. What's the total amount of solution in the tank after t minutes? The brine solution is coming in with the speed of 5 liters per minute. It is escaping out of the tank with speed of 3 minutes per minute. So for every minute, 2 liters of brine solution is the increasing. So after t minutes, the total amount of solution in the tank will be? Initially we have 100 litres and every minute we have 2 litres more, so after t minutes, we have 100 plus 2 times t litres of solution inside the tank. There we have, in total, m kilograms of salt, so m over 100 + 2 times t kilograms per liter is the concentration of salt inside of the tank after t minutes. This is the concentration of salt inside of the tank after t minutes. So multiply this too. Is coming out of the tank with a speed of 3 liters per minute, having concentration this much, so the total amount of salt which is coming out of the tank will be 3m over 100 plus 2t kilograms per minute. That's the output rate of salts. So now, the total amount of salt after t minutes in the tank must satisfy the initial value problem, m prime is equal to, input rate is this one, 1 minus, output rate is this one, so 1 minus 3m over 100 plus 2t and inicial condition, m(0) that is equal to ten. This is a simple initial variable problem. And this is a linear equation, in fact, so it is easy enough to solve. So this linear differential equation, you will get m(t) is equal to, finally, 1/5 times 2t + 100- 10 to the 4 times 2t + 100 to the -3/2 kilograms. That's the total amount of salt remaining in the tank after t minutes of experiment.