(problem 1) Find all solutions of the differential equation .
The solutions are: for any constant .
We solve separable differential equations and initial value problems.
Differential Equations
A differential equation is an equation that involves one or more derivatives of an unknown function. Solving a differential equation entails determining the unknown function.
example 1 Solve the differential equation .
We seek such that . By inspection is a solution of this equation. However, is also a solution. In fact, there are infinitely many solutions: , where is any constant (including 0). This family of solutions to the differential equation is called the general solution.
We seek such that . By inspection is a solution of this equation. However, is also a solution. In fact, there are infinitely many solutions: , where is any constant (including 0). This family of solutions to the differential equation is called the general solution.
An initial value problem is a differential equation along with other information about the solution, usually the value of the function at a point. The purpose of the initial value is to determine one specific solution of the differential equation, in the event that there was more than one solution.
example 2 Solve the initial value problem .
We saw in the last example that the differential equation has infinitely many solutions, , where is any constant. The initial value can be used to determine the value of so that we will have a unique solution. Since , we can see that . Thus the initial value problem has the unique solution .
We saw in the last example that the differential equation has infinitely many solutions, , where is any constant. The initial value can be used to determine the value of so that we will have a unique solution. Since , we can see that . Thus the initial value problem has the unique solution .
(problem 2) Solve the initial value problem .
The general solution of the differential equation is
The solution to the initial value problem is
Separable Differential Equations
A separable differential equation is a differential equation that can be put in the form . To solve such an equation, we separate the variables by moving the ’s to one side and the ’s to the other, then integrate both sides with respect to and solve for . In general, the process goes as follows: Let for convenience and we have
Integrating both sides with respect to using a the substitution , we have and so Now we solve for algebraically to get the final answer. To simplify the computations, we will use instead of and then make a slight abuse of notation to get to the same end result. From the beginning: again using for convenience. Now we write this last equation in differential form and put an integral sign on both sides(!): yielding as before.
example 3 Find the general solution of the differential equation and then find the
particular solution corresponding to the initial condition .
Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that has been replaced by , as this makes no difference. Now take reciprocals: Note that was replaced by as this makes no difference in the general solution. We use the initial condition to find the particular solution. Substitute and into the general solution to get: which simplifies to and we conclude by inspection that . The particular solution is thus Note that we can check the general solution to the differential equation by calculating and observing that . From the quotient rule, we have
Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that has been replaced by , as this makes no difference. Now take reciprocals: Note that was replaced by as this makes no difference in the general solution. We use the initial condition to find the particular solution. Substitute and into the general solution to get: which simplifies to and we conclude by inspection that . The particular solution is thus Note that we can check the general solution to the differential equation by calculating and observing that . From the quotient rule, we have
(problem 3a) Find the general solution of the differential equation
Separating the variables gives
Integrating gives
The general solution is
(problem 3b) Solve the initial value problem
Separating the variables gives
Integrating gives
The general solution is
The solution to the initial value problem is
example 4 Solve the differential equation .
Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that is a positive constant. Remove the absolute values: where is any non-zero constant. Looking back at the original differential equation, we see that is a solution, so we can say that is actually any constant. We can check the differential equation by calculating and observing that . From the chain rule, we have
Rewrite the equation as Move the to the left and the to the right to get Integrate both sides: This gives Solve for : Note that is a positive constant. Remove the absolute values: where is any non-zero constant. Looking back at the original differential equation, we see that is a solution, so we can say that is actually any constant. We can check the differential equation by calculating and observing that . From the chain rule, we have
(problem 4a) Find the general solution of the differential equation
Separating the variables gives
Integrating gives
The general solution is
(problem 4b) Solve the initial value problem
Separating the variables gives
Integrating gives
The general solution is
The solution to the initial value problem is
example 5 A population of 2 million bacteria doubles in 1 hour. A natural
assumption is that the population is growing at a rate which is proportional to the
number of bacteria present. Use this information to write a differential equation that
models the population, , at time . Solve the differential equation and find the
population after 90 minutes.
Let denote the population of the bacteria (in millions) after hours. Since the rate of growth is proportional to the number present, we have for some constant . Moving to the left and to the right and integrating gives Anti-differentiating gives: Solving for : Removing the absolute value bars, we have where is a nonzero constant. To determine the values of and we will use the given information about the size of the population at time and . The population was initially 2 million, so and in one hour it doubled, so . the first of these gives, Using the second condition gives: Thus is the formula for the population at time . Finally, after 90 minutes (or 1.5 hours), the population is million bacteria.
Let denote the population of the bacteria (in millions) after hours. Since the rate of growth is proportional to the number present, we have for some constant . Moving to the left and to the right and integrating gives Anti-differentiating gives: Solving for : Removing the absolute value bars, we have where is a nonzero constant. To determine the values of and we will use the given information about the size of the population at time and . The population was initially 2 million, so and in one hour it doubled, so . the first of these gives, Using the second condition gives: Thus is the formula for the population at time . Finally, after 90 minutes (or 1.5 hours), the population is million bacteria.
(problem 5a) In the year 2000, the human population on earth was approximately 6
billion people and in the year 2010 it was approximately 6.5 billion. If we assume that
the population is growing at a rate which is proportional to the number of people
present, write a differential equation that models the population at time and find
the population in the year 2050.
Let denote the population (in billions) years after the year 2000.
The differential equation which models human population is
for some constant .
The general solution to this differential equation is
The value of is .
The value of to 3 decimal places is .
The population in the year 2050 is projected to be
8.5 billion 9 billion 9.5 billion
(problem 5b) With continuous compounding of interest on money in a bank account,
the rate of growth is proportional to the amount present. How long will it take for
money in an account to double if the account earns a yearly interest rate with
continuous compounding?
Let denote the amount of money in the account at time in years.
The differential equation which models the amount in the account is
The general solution to this differential equation is
To determine the doubling time, we should let
The doubling time is approximately
9 years 9.5 years 10 years
example 6 A water tank contains 100 gallons of pure water. Salt water with a
concentration of 0.2 lbs/gal is added to the tank at a rate of 3 gallons per minute. At
the same time, water is drained from the tank at the same rate of 3 gallons
per minute. Find an expression for the amount of salt in the tank at time
.
Let represent the amount of salt in the tank at time . Then represents the rate of change in the amount of salt at time . Salt is entering the tank at a rate of 0.6 lb/min and salt is leaving the tank at a rate of 0.03A lbs/min. This leads to the differential equation To solve this differential equation, we first rewrite the derivative on the left side: Separating variables gives Integrate both sides: Computing the anti-derivatives, we have Solving for , we obtain Removing the absolute value bars, we have where K is a nonzero constant. subtracting and dividing by we have We can find the value of since we know that at time there was no salt in the tank, so that : Since we can write the amount of salt in the tank as Note that as , pounds of salt. This makes sense since we are adding salt at a concentration of 0.2 lbs/gal to a 100 gallon tank and 0.2 lbs/gal 100 gal = 20 lbs.
Let represent the amount of salt in the tank at time . Then represents the rate of change in the amount of salt at time . Salt is entering the tank at a rate of 0.6 lb/min and salt is leaving the tank at a rate of 0.03A lbs/min. This leads to the differential equation To solve this differential equation, we first rewrite the derivative on the left side: Separating variables gives Integrate both sides: Computing the anti-derivatives, we have Solving for , we obtain Removing the absolute value bars, we have where K is a nonzero constant. subtracting and dividing by we have We can find the value of since we know that at time there was no salt in the tank, so that : Since we can write the amount of salt in the tank as Note that as , pounds of salt. This makes sense since we are adding salt at a concentration of 0.2 lbs/gal to a 100 gallon tank and 0.2 lbs/gal 100 gal = 20 lbs.
(problem 6a) A water tank contains 60 gallons of salt water with 1 pound of salt
mixed in. Salt water with a concentration of 0.1 lbs/gal is added to the tank at a rate
of 1.5 gallons per minute. At the same time, water is drained from the tank at the
same rate of 1.5 gallons per minute. How long will it take for the tank to contain 2
pounds of salt?
Let represent the amount of salt in the tank at time minutes.
Salt is entering the tank at a rate of lb/min
Salt is leaving the tank at a rate of lbs/min
This leads to the differential equation Separating variables gives
Integrating gives
Solving for , we have
The value of is
The time it takes for the tank to contain 2 pounds of salt is approximately
9 minutes 9.5 minutes 10 minutes
(problem 6b) Newtons Law of Cooling states that the rate of cooling of an
object at temperature is proportional to the difference between and the
ambient temperature. A loaf of banana nut bread (with chocolate chips) is
removed from an oven at and placed on a cooling rack in a kitchen whose
temperature is . Two minutes later, the temperature of the bread is measured to be
. How much longer will take to be at the desired eating temperature of
The differential equation that models Newton’s Law of Cooling is
Separating variables gives
Integrating gives
Solving for , we have
The value of is .
The value of the constant of proportionality, is
-0.303 0.303 -0.606
The additional time until the bread is at is
2.7 minutes 4.7 minutes 6.7 minutes