Setting up and Solving Differential Equations

We have seen how to solve differential equations by the method of separating the variables. We have also met equations that describe situations of growth and decay. This presentation brings the 2 topics together and we see how to set up and solve the differential equations for growth and decay. We will also set up and solve some differential equations that describe other situations.

e. g. 1. A solution initially contains 200 bacteria e.g. 1. A solution initially contains 200 bacteria. Assuming the number, x, increases at a rate proportional to the number present, write down a differential equation connecting x and the time, t. If the rate of increase of the number is initially 100 per hour, how many are there after 2 hours? Solution: The description in the question is typical of exponential growth. We have to set up a differential equation which describes the situation and solve it to find x when t = 2.

e. g. 1. A solution initially contains 200 bacteria e.g. 1. A solution initially contains 200 bacteria. Assuming the number, x, increases at a rate proportional to the number present, write down a differential equation connecting x and the time, t. If the rate of increase of the number is initially 100 per hour, how many are there after 2 hours? Solution: The description in the question is typical of exponential growth. We have to set up a differential equation which describes the situation and solve it to find x when t = 2. , where k is a constant.

You may remember the solution to this equation but, if not, we can separate the variables to find it. We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour” There are 2 pairs of conditions here which enable us to solve for 2 unknowns.

You may remember the solution to this equation but, if not, we can separate the variables to find it. We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour”

You may remember the solution to this equation but, if not, we can separate the variables to find it. We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour”

You may remember the solution to this equation but, if not, we can separate the variables to find it. We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour” Substituting in (1):

You may remember the solution to this equation but, if not, we can separate the variables to find it. We were given “ A solution initially contains 200 bacteria . . . “ and “. . . the rate of increase of the number is initially 100 per hour” Substituting in (1):

The graph showing the growth function is Number after 2 hours Number at start of measurements

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days.

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We could write where k is negative, but most people prefer to emphasise the negative gradient.

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We can quote the solution:

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We can quote the solution: Make sure you write t on the r.h.s. !

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We can quote the solution:

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We can quote the solution: A log is just an index !

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. Solution: Let m be mass in mg and t time in days. where k is a positive constant. We can quote the solution:

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. We now have

e.g. 2 A radioactive element decays at a rate that is proportional to the mass remaining. Initially the mass is 10 mg and after 20 days it is 5 mg. Set up a differential equation describing this situation and solve it to find the time taken to reach 1 mg. We now have ( nearest integer ) It takes 66 days to decay to 1 mg.

The graph showing the decay function is Mass at start of measurements Time when mass is 1 mg

SUMMARY The words “ a rate proportional to . . . ” followed by the quantity the rate refers to, gives the differential equation for growth or decay. e.g. “ the number, x, increases at a rate proportional to x ” gives The solution to the above equation is ( but if we forget it, we can easily separate the variables in the differential equation and solve ). The values of A and k are found by substituting either 1 pair of values of x and t and 1 pair of values of and t, or 2 pairs of values of x and t.

Exercise For the following problems, choose suitable letters and set up the differential equations but don’t solve them. When you have the first 2 equations, check you agree with me and then solve the complete problems.

Exercise 1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred. ( When solving, use k correct to 3 s.f. ) 2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute. ( When solving, use k correct to 3 s.f. )

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred. Solution: Let t = 0 in 1990.

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred. ( nearest hundred )

2. A patient is receiving drug treatment 2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute. Solution: (3 s.f.)

2. A patient is receiving drug treatment 2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute. Substitute Ans: 5 hrs 44 mins

You might meet differential equations that do not describe growth functions.

e.g. 1 The gradient of a curve at every point equals the square of the y-value at that point. Express this as a differential equation and find the particular solution which passes through ( 1, 1 ). Solution: The equation is

e.g. 1 The gradient of a curve at every point equals the square of the y-value at that point. Express this as a differential equation and find the particular solution which passes through ( 1, 1 ). Solution: The equation is Separating the variables: This is the general solution to the equation

( 1, 1 ) lies on the curve So, or, The equation is

Exercise 1. The gradient of a curve at any point ( x, y ) is equal to the product of x and y. The curve passes through the point ( 1, 1 ). Form a differential equation and solve it to find the equation of the curve. Give your answer in the form . Solution:

( 1, 1 ) on the curve: So,

There is one very well known situation which can be described by a differential equation. The following is an example.

The temperature of a cup of coffee is given by at time t minutes after it was poured. The temperature of the room in which the cup is placed is Explain what the following equation is describing: Solution: The equation gives the rate of decrease of the temperature of the coffee. It is proportional to the amount that the temperature is above room temperature. This is an example of Newton’s law of cooling.

We can solve this equation as follows: If we are given further information, we can complete the solution as in the other examples.