1. Chapter 9.3: Modeling with First-Order Differential Equations
Unit 8 Lesson 5
Chapter 9.3: Modeling with First-Order Differential Equations

2. Review Solve the differential equation, 𝒅𝒚 𝒅𝒙 =𝟒−𝒙 𝑑𝑦 𝑑𝑥 =4−𝑥
𝑑𝑥 𝑑𝑦 𝑑𝑥 =(4−𝑥) 𝑑𝑥
𝑑𝑦= 4−𝑥 𝑑𝑥
𝑑𝑦 = 4−𝑥 𝑑𝑥
𝑦=4𝑥− 𝑥 𝐶

3. Algebra Review: Directly Proportional
Two quantities are said to be in direct proportion (or directly proportional, or simply proportional), if one quantity is a constant multiple of the other. For example, if y is proportional to x and if k is a constant then:
𝑘= 𝑦 𝑥
𝑦=𝑘∙𝑥 or

4. Algebra Review: Inversely Proportional
Two quantities are said to be in inverse proportion if their product is constant. ( In other words, when one variable increases the other decreases in proportion so that the product is unchanged.) For example, if y is inversely proportional to x and if k is a constant, then:
𝑦= 𝑘 𝑥 or
𝑦∙𝑥=𝑘

5. Algebra Review: Jointly Proportional
When we say that z is jointly proportional to a set of variables, it means that z is directly proportional to each variable taken one at a time. For example, if z is proportional to x and to y, and if k is a constant, then:
𝑧=𝑘𝑥𝑦

6. Creating proportional formulas using differential equations
The rate of change of P with respect to t is proportional to 10 – t. Find the function.
𝑑𝑃 𝑑𝑡 =𝑘 10−𝑡
Direct proportion!
𝑦=𝑘𝑥
𝑑𝑃=𝑘 10−𝑡 𝑑𝑡
“rate of change of P with respect to t” is code for
𝑑𝑃 =𝑘 10−𝑡 𝑑𝑡
𝑑𝑃 𝑑𝑡
𝑃=𝑘 10𝑡− 1 2 𝑡 2 +𝐶

7. 𝑑𝑎 𝑑𝑏 Your Turn 𝑑𝑎 𝑑𝑏 = 𝑘 𝑏 = 𝑘 𝑏 1 2 =𝑘 𝑏 − 1 2 𝑦= 𝑘 𝑥
The rate of change of a with respect to b is inversely proportional to the square root of b. Find the function.
Inverse proportion!
𝑑𝑎 𝑑𝑏 = 𝑘 𝑏 = 𝑘 𝑏 =𝑘 𝑏 − 1 2
𝑦= 𝑘 𝑥
“rate of change of a with respect to b” is code for
𝑑𝑎 =𝑘 𝑏 − 1 2 𝑑𝑏
𝑑𝑎 𝑑𝑏
𝑎=𝑘 2 𝑏 𝐶

8. Growth and Decay
In many applications, the rate of change of a variable is proportional to the value of y
If 𝑑𝑦 𝑑𝑡 =𝑘𝑦, then the equation y = Cekt is the general solution
C is the initial value of y
k is the proportional constant
Exponential Growth occurs when k > 0
Exponential Decay occurs when k < 0
In Precalc, we used the formula 𝐴=𝑃 𝑒 𝑟𝑡 . Is there a difference?

9. Deriving the formula 𝑦=𝐶 𝑒 𝑘𝑡
The derivative IS written as a product of a function of t and a function of y.
Find the general solution to:
Now use separation of variables to find the general solution.
Since C is arbitrary, ±eC represents an arbitrary nonzero number. We can replace it with C:

10. Continuous Growth and Decay
If: Then:
The rate of change in the output (dy/dt) is proportional to the output (y).
Your Choice: Remember how to derive the general solution from the differential equation OR memorize the general solution.

11. Example
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
Write a differential equation that states this fact. Evaluate the proportionality constant if, at the time of zero, the pressure is 35 psi and decreasing at a rate of .28 psi/min.
Solve the differential equation subject to the initial condition in a.
Sketch the graph of the function without a calculator. Show its behavior a long time after the tire is punctured.
What would the pressure be 10 minutes after the tire was punctured?
The car is safe to drive as long as the tire pressure is 12 psi or greater. For how long after the puncture will the car be safe to drive?

12. Example: Part A 𝑑𝑃 𝑑𝑡 =𝑘𝑃 −0.28=𝑘(35) 𝑘= −0.28 35 =−0.008
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
Write a differential equation that states this fact. Evaluate the proportionality constant if, at the time of zero, the pressure is 35 psi and decreasing at a rate of .28 psi/min.
Why is 𝑑𝑃 𝑑𝑡 negative here?
𝑑𝑃 𝑑𝑡 =𝑘𝑃
−0.28=𝑘(35)
This is the diff eqn
𝑘= − =−0.008
In this case, P represents pressure!
the proportionality constant

13. Example: Part B From Part A 𝑑𝑃 𝑑𝑡 =−0.008𝑃 𝑒 ln 𝑃 = 𝑒 −0.008𝑡+𝐶
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
B. Solve the differential equation subject to the initial condition in a. (P = 35 psi)
𝑡,𝑃 →(0,35)
From Part A
𝑑𝑃 𝑑𝑡 =−0.008𝑃
𝑒 ln 𝑃 = 𝑒 −0.008𝑡+𝐶
𝑑𝑃 𝑑𝑡 =𝑘𝑃
𝑃= 𝑒 −0.008𝑡 ∙ 𝑒 𝑐
1 𝑃 𝑑𝑃= −0.008 𝑑𝑡
𝑃=𝐶 𝑒 −0.008𝑡
𝑘=−0.008
1 𝑃 𝑑𝑃 =− 𝑑𝑡
35=𝐶 𝑒 −0.008(0)
𝐶=35
ln 𝑃 =−0.008𝑡+𝐶
𝑃=35 𝑒 −0.008𝑡

14. Example: Part C
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
C. Sketch the graph of the function without a calculator. Show its behavior of the graph over an extended period of time after the tire is punctured.

15. Example: Part D 𝑃 10 =35 𝑒 −0.008(10) 𝑃 10 =32.309 𝑝𝑠𝑖 𝑃=35 𝑒 −0.008𝑡
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
D. What would the pressure be 10 minutes after the tire was punctured?
𝑃=35 𝑒 −0.008𝑡
𝑃 10 =35 𝑒 −0.008(10)
𝑃 10 = 𝑝𝑠𝑖

16. Example: Part E ln 12 35 −0.008 =𝑡 ln 12 35 = ln 𝑒 −0.008𝑡
Your tires have just run over a nail. As air leaks out of the car’s tire, the rate of change of the air pressure inside the tire is directly proportional to the pressure.
E. The car is safe to drive as long as the tire pressure is 12 psi or greater. For how long after the puncture will the car be safe to drive?
ln −0.008 =𝑡
ln = ln 𝑒 −0.008𝑡
𝑃=35 𝑒 −0.008𝑡
12=35 𝑒 −0.008𝑡
𝑡= 𝑚𝑖𝑛
ln =−0.008𝑡
12 35 = 𝑒 −0.008𝑡
𝑡= 𝑚𝑖𝑛× 1 ℎ𝑟 60 𝑚𝑖𝑛
If you need to convert to hours, divide by 60
𝑡≈2.230≈2 ℎ𝑟𝑠 14 𝑚𝑖𝑛

17. Example 2 𝑡,𝐵 → 0,500 1000=500 𝑒 𝑘(2) 500=𝐶 𝑒 𝑘(0) 2,1000 𝐶=500
A certain type of bacteria increases continuously at a rate proportional to the number present. If there are 500 present at a given time and present 2 hours later, how many will there be 5 hours from the initial time given?
Problem! We don’t know C or k! Solving for C is easy!
Solving for k is trickier.
𝑡,𝐵
→ 0,500
500=𝐶 𝑒 𝑘(0)
1000=500 𝑒 𝑘(2)
2,1000
2 hrs later
𝐶=500
2= 𝑒 2𝑘
5,???
5 hrs later
𝐵=500 𝑒 𝑘𝑡
ln 2 = ln 𝑒 2𝑘
Since we know growth is proportional, we can say …
ln 2 =2𝑘
𝑑𝐵 𝑑𝑡 =𝑘𝐵
𝐵=500 𝑒 ln 2 2 (𝑡)
→𝐵=𝐶 𝑒 𝑘𝑡
𝑘= ln 2 2
Now we are ready to answer the question!

18. Example 2 (cont) 𝐵=500 𝑒 ln 2 2 (𝑡) 𝑡,𝐵 → 0,500 2,1000
A certain type of bacteria increases continuously at a rate proportional to the number present. If there are 500 present at a given time and present 2 hours later, how many will there be 5 hours from the initial time given?
𝐵=500 𝑒 ln 2 2 (𝑡)
𝑡,𝐵
→ 0,500
2,1000
2 hrs later
𝐵 5 =500 𝑒 ln 2 2 (5)
5,???
5 hrs later
𝐵 5 = ≈2828 𝑏𝑎𝑐𝑡𝑒𝑟𝑖𝑎

19. Example 3 (see if you can do it!)
The rate of change of 𝑁′ is proportional to N. When 𝑡=0, 𝑁=250 and when 𝑡=1, 𝑁= What is the value of 𝑁 when 𝑡=4?
From experience, we know that when t = 0, then 𝐶= 𝑁 0
Solving for k
Set up relationships
400=250 𝑒 𝑘(1)
𝑡,𝑁
→ 0,250
𝑁=250 𝑒 𝑘𝑡
𝑡=1
1,400
8 5 = 𝑒 𝑘
Put it all together and answer the orig question
𝑡=4
4,???
ln 8 5 = ln 𝑒 𝑘
𝑁=250 𝑒 ln 𝑡
𝑑𝑁 𝑑𝑡 =𝑘𝑁
𝑁 4 =250 𝑒 ln
𝑘= ln 8 5
→𝑁=𝐶 𝑒 𝑘𝑡
𝑁 4 =1638.4

20. Example 4 300= 100 𝑒 2𝑘 𝑒 4𝑘 𝑡,𝐹 → 0,??? 𝐹=𝐶 𝑒 𝑘𝑡 𝑡=2 2,100 100=𝐶 𝑒 2𝑘
Suppose an experimental population of fruit flies increases according to the law of exponential growth. There were 100 flies after the second day of the experiment and 300 flies after the fourth day. Approximately how many flies were in the original population?
Set up relationships
C is not going to be pretty (and temporary)
Solving for k with funky C
300= 𝑒 2𝑘 𝑒 4𝑘
𝑡,𝐹
→ 0,???
𝐹=𝐶 𝑒 𝑘𝑡
𝑡=2
2,100
100=𝐶 𝑒 2𝑘
𝑡=4
4,300
300=100 𝑒 2𝑘
𝐶= 100 𝑒 2𝑘
𝑑𝐹 𝑑𝑡 =𝑘𝐹
3= 𝑒 2𝑘
𝑘= ln 3 2
ln 3 = ln 𝑒 2𝑘
→𝐹=𝐶 𝑒 𝑘𝑡
ln 3 =2𝑘
continue…

21. Example 4 (cont) 𝑒 2 ln 3 2 =3 𝐹= 100 3 𝑒 ln 3 2 𝑡 𝑘= ln 3 2 𝑡,𝐹
Suppose an experimental population of fruit flies increases according to the law of exponential growth. There were 100 flies after the second day of the experiment and 300 flies after the fourth day. Approximately how many flies were in the original population?
Set up relationships
𝐹= 𝑒 ln 𝑡
𝑘= ln 3 2
𝑡,𝐹
→ 0,???
Figuring out # of flies at t=0
Let’s clean up that C!
𝑡=2
2,100
𝐹 0 = 𝑒 ln (0)
𝐶= 100 𝑒 2𝑘
𝑡=4
4,300
𝑑𝐹 𝑑𝑡 =𝑘𝐹
𝐹 0 = =33.333
𝐶= 100 3
→𝐹=𝐶 𝑒 𝑘𝑡
𝐹 0 ≈33 𝑓𝑙𝑖𝑒𝑠

22. 𝑦= 𝑦 0 𝑒 −𝑘𝑡 Radioactive Decay
The equation for the amount of a radioactive element left after time t is:
𝑦= 𝑦 0 𝑒 −𝑘𝑡
𝑦 represents the amount after a specified period of time (𝑡)
𝑦 0 represents the initial amount of material

23. The half-life is the TIME required for half the material to decay.
That is, the time it takes for 𝑦= 1 2 𝑦 0
𝑦= 𝑦 0 𝑒 −𝑘𝑡
ln = ln 𝑒 −𝑘𝑡
Half-life:
1 2 𝑦 0 = 𝑦 0 𝑒 −𝑘𝑡
ln 1 − ln 2 =−𝑘𝑡
1 2 = 𝑒 −𝑘𝑡
ln 2 =𝑘𝑡
𝑡= ln 2 𝑘

24. Newton’s Law of Cooling
Newton's Law of Cooling states that the rate of change of the temperature of an object with respect to time (𝑡) is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).
Differential Equation
𝑑𝑇 𝑑𝑡 =−𝑘 𝑇 0 − 𝑇 𝑠
Temperature of object’s surroundings
Temperature at given time 𝑡
Solution
𝑇= 𝑇 𝑠 + 𝑇 0 − 𝑇 𝑠 𝑒 −𝑘𝑡
Initial temperature of object

25. Example using Newton’s Law of Cooling
A cup of water with a temperature of 95°𝐶 is placed in a room with a constant temperature of 21°𝐶.
A) Assuming Newton’s Law of Cooling applies, set up and solve an initial-value problem whose solution is the temperature of the water 𝑡 minutes after it is placed in the room.
ln 𝑇−21 =−𝑘𝑡+ 𝐶 1
𝑑𝑇 𝑑𝑡 =−𝑘(𝑇−21)
𝑇−21=𝐶 𝑒 −𝑘𝑡
95=21+𝐶 𝑒 −𝑘(0)
1 𝑇−21 𝑑𝑇=−𝑘𝑑𝑡
𝑇=21+𝐶 𝑒 −𝑘𝑡
𝐶=74
1 𝑇−21 𝑑𝑡 =−𝑘 𝑑𝑡
𝑇=21+74 𝑒 −𝑘𝑡

26. Example using Newton’s Law of Cooling
A cup of water with a temperature of 95°𝐶 is placed in a room with a constant temperature of 21°𝐶.
B) How many minutes will it take for the water to reach a temperature of 51°𝐶 if it cools to 85°𝐶 in 1 minute?
𝑇=21+74 𝑒 −𝑘𝑡
𝑇=21+74 𝑒 ln 𝑡
51=21+74 𝑒 ln 𝑡
85=21+74 𝑒 −𝑘(1)
64=74 𝑒 −𝑘(1)
30=74 𝑒 ln 𝑡
32 37 = 𝑒 −𝑘(1)
15 37 = 𝑒 ln 𝑡
ln = ln 𝑡
𝑘=− ln
𝑡=6.219 𝑚𝑖𝑛

27. Assignment
Worksheet