Differential equations arise in many areas of science and technology. This lecture will show examples where differential equations can be widely applied to model natural phenomena, engineering systems and many other situations.
Newton's Law of Cooling
Example 1:
Problem:
A cup of coffee is initially 170 degrees Fahrenheit and is left in a room with ambient temperature 70 degrees Fahrenheit. Suppose that when the coffee is first placed in the room, it is cooling at a rate of 20 degrees per minute.
Assuming Newton's law of cooling applies, how long does it take for the coffee to cool to 110 degrees?
Solution:
Newton's law of cooling states that this differential equation holds for the temperature of a cooling object:
where
T = temperature of object
t = time (in minutes for this problem)
A = ambient temperature
k = constant
We know all these terms except for k, so we set up the equation and solve for k.
At time t = 0, we are told that
so now we know
Next we solve the differential equation for T using separation of variables.
Again, the initial conditions let us solve for C.
Then, substitute 110 for the temperature and solve for t.
Therefore, the coffee cools to the desired temperature in about 4.581 minutes.
Population
Example 2:
Problem:
Suppose the growth of a population is proportional to the population itself. If the population of a colony doubles in 50 days, in how, many days will the population become triple.
Solution:
Let the initial population be P0 and P be the population of the colony at any instant t.
Then according to the problem
log P = kt + log P0 ….(2)
when P0 is doubled,P = 2P0 where t = 50 days
From equation (2)log (2 P0) = 50 k + log P0
Mixtures
Example 3:
Problem:Example 3:
We have a 100-gallon tank filled with water. Saltwater is being pumped into the tank through two pipes. One pipe adds water at a rate of 3 gallons per minute, and the salt concentration is 12 grams per gallon. The other pipe adds water at a rate of 2 gallons per minute, with salt concentration of 7 grams per gallon. Water drains out of the tank through a hole in the bottom at the rate of 5 gallons per minute, so the total amount of water in the tank is kept constant at 100 gallons.
Assume that the tank is kept well-mixed, so that at any point in the tank, the concentration of salt is the same as at any other point. If the amount of salt initially in the tank is 20 grams,
1. Write an equation which states how the amount of salt S is changing with respect to time.
2. Use the differential equation from part 1 to write a function that states how much salt is present in the tank at a given time t.
Solution:
1. Finding the differential equation
The rate of change of the amount of salt depends on how quickly salt is entering the tank through the pipes, and how quickly it is leaving through the hole. Thus, the equation has the form
The amount of salt entering through the first pipe (per minute) is
and through the second pipe
The amount of salt leaving through the hole is trickier to compute, since it depends on the concentration of the salt in the water. Since the tank is well-mixed, if there are S grams of salt in the 100-gallon tank, the concentration is
and so the amount of salt leaving per minute is
Therefore
2. Solving the differential equation
To find the function that states how much salt is present in the tank at a given time t, we now solve the differential equation
by separation of variables:
Integrating both sides, we get
where C is an arbitrary constant. Simplify and solve for S:
Next, solve for the constant C4 by using the initial condition t = 0, S = 20:
Therefore, the equation giving the amount of salt in the tank at time t is
Sources:
math.dartmouth.edu
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