Math 1526:
A differential equation is an equation that relates not only a dependent variable and an independent variable, but also one or more derivatives of that dependent variable. For example, x + y + y ´ = 0 is a differential equation. After completing this project, you should be familiar with basic terminology related to differential equations, and should also be able to solve some simple differential equations.
Why are differential equations useful?
Differential equations allow us to model situations in which the rate of change of some quantity depends directly on the quantity itself. Here are some examples that should illustrate the difference:

When you drive your car, you have the ability to independently choose your speed. The speed of the car doesn’t depend on how far the car has already traveled. Therefore, we would not want to use a differential equation to model this.

When you inflate a balloon, there is a probability that the balloon will get too full and burst. That probability changes as you add air, and that rate of change depends on how much air you have already added previously. Therefore, we would want to use a differential equation to model this.

You invest $10,000 in a long term, interest-bearing fund. Since your interest is compounded, the amount of interest you make in future years depends on how much interest you have already made. Therefore, we could use a differential equation to model this. (See Example 4.)
Next we’re going to look at some examples of differential equations.
What is a differential equation, and what does it mean to ‘solve’ one?
Consider the equation y
´´ – 2 y
´ = 3 y . This is a differential equation, because some of the variables are derivatives of other variables. In order for a function to solve a differential equation, the equation must be true when the function and its correct derivatives are substituted into the equation.

The function y = e
3 x
is a solution to the differential equation y
´´ – 2 y
´ = 3 y : y
´ = 3e 3 x y
´´ = 9e 3 x
Substituting these into the original equation gives us:
9e
3 x
– 2(3e
3 x
) = 3(e
3 x
)
9e
3 x
– 6e
3 x
= 3e
3 x
3e
3 x
= 3e
3 x
Since the left-hand side equals the right-hand side, the function y = e
3 x is a solution.

The function y = x
2
is not a solution to the differential equation y
´´ – 2 y
´ = 3 y : y
´ = 2 x y
´´ = 2

Substituting these into the original equation gives us:
2 – 2(2 x ) = 3( x 2 )
2 – 4 x = 3 x
2
Technically that last equation is true for some isolated values of x , ( x ≈ 0.3874, for example), but it’s not generally a true statement like 3e 3 x
= 3e
3 x
obviously is. Therefore, we determine that y = x
2
is not a solution to this differential equation.
What do we mean when we talk about the ‘order’ of a differential equation?
The order of a differential equation is the highest derivative that appears within the differential equation. For example, y + 8 x
2
= 3 xy
´ + y
´´ – y
´´´ is a third-order differential equation (because it contains a y ´´´ term.)
For the remainder of this project, we focus exclusively on first-order differential equations that can be solved using a technique called separation of variables .
How do we solve a first-order differential equation?
In this example, we’ll solve the differential equation y
 
3 x
2
. Solving differential equations y like this one requires four steps:
Step one: Rewrite the derivative y
´ as dy dx dy
. This gives us dx

3 x
2 y
.
Step two: Rearrange the terms of the equation so that all the terms containing y are on the lefthand side and all the terms containing x are on the right-hand side. dy dx

3 x
2 y y dy

3 x
2 dx
(Note: It is not always possible to separate the y -terms from the x -terms like this. If that isn’t possible, the equation is called a non-separable differential equation. Solving non-separable equations is beyond the scope of this course.)
Step three:
Find the antiderivative of both sides. Be sure to include “+
C
” on the right-hand side.
 y
2
2
Step four: Solve for y . y dy



3 x
2 x
3 
C dx
Technically, we get a constant on both sides, and then combine them into the one constant you see here.

y
2
 x
3 
C
2 y
2 
2 x
3 
C y
 
2 x
3 
C
So, any function in the form y
 
2 x
3 
C will be a
Q: When we multiplied by 2, why didn’t the
C change to a 2 C ?
A:
There’s nothing wrong with writing 2 C here, but 2 times an unknown value is still an unknown value! Some people will use a new letter here, such as K , to show that the value of the constant has changed. solution to the differential equation.
How about a word problem?
Compound interest means that the rate of change in your balance is directly proportional to the balance. That is, if you invested $10,000 at 6% interest, then dA

0 .
06 A . (Literally, that dt equation reads “The rate of change in your balance over time, is equal to 0.06 — in other words,
6% — times your current balance.”)
Solve this differential equation. Note that “Step one” from the previous example has already been done for us.
Step two: Rearrange the terms of the equation so that all the terms containing A are on the left and all the terms containing t are on the right. dA

0 .
06 A dt
1
A dA

0 .
06 dt
Step three: Find the antiderivative of both sides. Be sure to include “+ C ” on the right-hand side.

Step four: Solve for A .
1
A dA ln

A


0 .
06
0 .
06 t
 dt
C
The antiderivative of ln
1
A
is
A . But since we know that balance must be positive
(you can’t invest negative money!), we can safely omit the absolute value signs here.
ln A

0 .
06 t

C e ln A  e
0 .
06 t

C
A
 e
0 .
06 t

C

At this point, we could stop and say we were done, but our answer is not yet in simplest form.
There’s a special trick we can apply when +
C ends up in the exponent.
Step five (only when + C is in the exponent): Use the laws of exponents to put the constant out front.
A
A

 e e
0 .
06 t

C
0 .
06 t  e
C
If you’re rusty on the laws of exponents, the main point is that a + C in an exponent can be moved to a c in front of
A
 ce
0 .
06 t
(define c
 e
C
.) the term containing the exponent.
At this point, we’ve solved the equation. But in fact, we have enough information in this problem to figure out the value of c as well. Since our initial balance (when t = 0) was $10,000:
10000
 ce
0 .
06

0
10000
 c ( 1 ) c

10000
Therefore, A = 10000e
0.06
t . This shouldn’t be a surprising result, since we learned this formula back in 1525 for solving compound interest problems. Now you know where it comes from!
A pair of values that allows us to solve for c
—for example, the pair t = 0, A = 10,000 in the previous example—is called an initial value condition . Not all problems will have initial value conditions, however.
One more application: Newton’s Law of Cooling
Three hundred years ago, Isaac Newton discovered the Law of Cooling , which gives a formula for the rate at which objects cool down to room temperature when removed from an oven. This law can also be used in marketing applications, and here is an example:
A business normally enjoys sales of
$6000 per day. The owner decides
10
Example 5 to invest in a two-week series of television spots, and while those spots are running, her sales increase to $10000 per day. Three
9
8
7 days after the advertising campaign ends, however, her sales decline to
$8000 that day. What sales level
6
5 usual sales level should she expect seven days after the campaign ends?
0 2 4 6 days since end of ad campaign
8 10
Here’s one way to “visualize” the problem: Her usual sales were at one level: $6000 per day
(we’ll call this usual level S ). Advertising caused her sales to “heat up” to a higher level of
$10000 per day. Once the advertising ends, those sales start to “cool back down” to their usual

level (see the graph). Newton’s Law says that the rate of cooling over time is directly proportional to the difference between the current sales level and S . In other words, dy
  k ( y

S ) dt where k is the rate of cooling, y is the current (temporary) sales level, S is the usual sales level, and t is the number of days since the end of the promotion. Note that k and S are constants in this problem; t and y are our only variables.
Let’s solve this differential equation. Once again, “step one” has already been done for us:
Step two: Rearrange the terms of the equation so that all the terms containing y are on the left and all the terms containing t are on the right: dy
  k ( y

S ) dt
1 y

S dy
  k dt
Step three:
Find the antiderivative of both sides. Be sure to include “+
C
” on the right-hand side. ln( y – S ) is an antiderivative of
1 y

S
whenever constant. (You can check this by taking the derivative of the answer.)
S is a
 1 y

S ln( dy
 y

S )

  k dt
 kt

C
Be careful! Don’t write
 k
2
2 on this side. k is a constant in this problem!
Step four: Solve for y . ln( y

S )
  kt

C y

S
 e
 kt

C y

S
 e
 kt

C
Step five (only when + C is in the exponent): Use the laws of exponents to put the constant out front. y y


S
S
 e
 kt

C
 ce
 kt
Now we know what formula to use, y = S + ce
– kt
. Here are what the different letters in the formula stand for:


S represents her usual sales level, $6000 per day in this problem. Let’s choose
“thousands of dollars” as our unit—to make the numbers smaller—and set
S = 6.
 c represents the amount of boost that the advertising created. Since the advertising temporarily raised her sales to $10,000 per day, that’s a $4000 boost over her usual level, so c = 4.
 t and y are the variables, representing the number of days since the end of the advertising campaign, and the amount of sales that day, respectively.
 k represents the rate of cooling. We need to solve for k in order to finish this problem.
To solve for k , we’ll use the fact that on the third day ( t = 3), the sales were only $8000 ( y = 8).
You will always need to use an intermediate value to solve for k
—the starting or ending sales levels won’t work!
y

S
 ce
 kt
8

6

4 e
 k ( 3 )
2

4 e

3 k
1
 e

3 k ln
2
1
2
 
3 k k
 ln
1
2

3

0 .
231
Now that we know what k is, we can answer the original question: What sales should she expect on the seventh day after the campaign ends? y

S
 ce
 kt y

6

4 e

( 0 .
231 )( 7 ) y

6 .
794
Therefore, her sales on the seventh day should be approximately $6,794.


Work through the example problems in the introduction before attempting these problems.
Do not turn in the worked examples from the introduction. Also, do not hand in this document with your lab.
 Work all problems in order. Do not say “see attached” or other notation referring the grader to another location in the lab for part of a problem.

Label problems and parts of problems with appropriate numbers and/or letters.

You may write your answers neatly by hand for this project.
1. You are investing $50,000 in an interest-bearing account at 8% interest. Write a differential equation that models the balance in your account after t years ( Hint: See Example 4), and solve that equation completely.
2. Solve the following differential equations. You do not need to solve for C in these problems, but you do need to remove C from the exponent when appropriate. In part c , you have an initial value condition, which will allow you to solve for the constant. a. y '

2 x
3 c. y
´ = 9 x
2 y. y b. xy
´ = 3
For part c, the initial value condition is that when x = 2, y = 20.
3. A locally owned grocery store has an average sales level of $8000 per day. One week the owner announced that he would split that week’s profits 50-50 with the local school system, and sales that week skyrocketed to $20000 per day. However, four days after the promotion ended, sales were already back down to $13000 per day. What level of sales should the owner expect ten days after the end of the promotion? (Use thousands of dollars for your units, as in the example).