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CHAPTER 7 FirstOrder Differential Equations 7 7.1 MODELING WITH DIFFERENTIAL EQUATIONS 7.2 SEPARABLE DIFFERENTIAL EQUATIONS 7.3 DIRECTION FIELDS AND EULER’S METHOD 74 SYSTEMS OF FIRST ORDER DIFFERENTIAL 7.4 SYSTEMS OF FIRSTORDER DIFFERENTIAL EQUATIONS Slide 1 © The McGrawHill Companies, Inc. Permission required for reproduction or display.
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CHAPTER

7First‐Order Differential Equations 7q

7.1 MODELING WITH DIFFERENTIAL EQUATIONS7.2 SEPARABLE DIFFERENTIAL EQUATIONS7.3 DIRECTION FIELDS AND EULER’S METHOD7 4 SYSTEMS OF FIRST ORDER DIFFERENTIAL7.4 SYSTEMS OF FIRST‐ORDER DIFFERENTIAL 

EQUATIONS

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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS

Exponential GrowthThe table indicates the number of E. coli bacteria (in millions of bacteria per ml) in a laboratory culture measured at half‐hour intervals during the course of an experiment. 

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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS

Exponential GrowthThe plot appears to indicate that the bacterial culture is growing exponentially. Careful analysis of experimental data has shown that the rate at which the bacterial culture grows is directly proportional to the current population.

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Exponential GrowthLet y(t) represent the number of bacteria in a culture at time t, then the rate of change of the population withrespect to time is y’(t). Thus, since y’(t) is proportional to y(t), we have the following differential equation

where k is the growth constant. We wish to solve this equation for y(t). We begin by rewriting and integrating.

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Exponential GrowthWe may combine c1 and c2 as a single constant we lablel c. 

Since y(t) represent a population, y(t) > 0.

Solving for y(t) and defining A = ec we obtain the solution.

We call this the general solution of the differentialWe call this the general solution of the differential equation. For k > 0, it is called an exponential growth law and for k < 0 it is an exponential decay law

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and for k < 0, it is an exponential decay law.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.1 Exponential Growth in a Bacterial Colony

A freshly inoculated bacterial culture of Streptococcus Acontains 100 cells. When the culture is checked 60 minutes later, it is determined that there are 450 cells present. 

Assuming exponential growth, determine the number of cells present at any time t (measured in minutes) and find the doubling time.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.1 Exponential Growth in a Bacterial Colony

SolutionExponential growth means                                                        where A and k are constants to be determined.

We were given the initial condition . b i i hi i h l l i iSubstituting this into the general solution gives us 

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.1 Exponential Growth in a Bacterial Colony

SolutionWe can use the second observation to find k.

We now have a formula representing the number of cells present at any time t:present at any time t:

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.1 Exponential Growth in a Bacterial Colony

SolutionTo find the doubling time, we wish to find t such that

We substitute this into y(t) and solve for t.

We see that the population doubles about every 28 

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minutes.

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Exponential DecayExperiments have shown that the rate at which a radioactive element decays is directly proportional to the amount present. Let y(t) represent the amount (mass) of a radioactive element present at time t, then the rate of decay satisfies

We know the general solution to this differential equation.

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Exponential DecayIt is common to discuss the decay rate of a radioactive element in terms of its half‐life, the time required for half of the initial quantity to decay into other elements. 

For instance, scientists have calculated that the half‐life of carbon‐14 (14C) is approximately 5730 years. That is, if you 

14have 2 grams of 14C today and you come back in 5730 years, you will have approximately 1 gram of 14C remaining.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.2 Radioactive Decay

If you have 50 grams of 14C today, how much will be left in 100 years?

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.2 Radioactive Decay

SolutionLet y(t) be the mass (in grams) of 14C present at time t. Then, we haveThe initial condition y(0) = 50 gives 

Using the half‐life, we can find the decay constant k.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.2 Radioactive Decay

SolutionThe graph shows the incredibly slow decay of 14C.

If we start with 50 grams, then the amount left after 100 years is

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Newton’s Law of CoolingThe rate at which the object cools (or warms) is not proportional to its temperature, but rather, to the difference in temperature between the object and itssurroundings. 

Symbolically, if we let y(t) be the  temperature of the object at time t and let Ta be the temperature of the surroundings (the ambient temperature, which we assume to be constant), we have the differential equation

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Newton’s Law of CoolingW i h t l thi ti f (t) W b i bWe wish to solve this equation for y(t). We begin by rewriting and integrating.

To evaluate the integral, we will use the substitution               u = y(t) − T and we assume y(t) − T > 0u = y(t)   Ta and we assume y(t)   Ta > 0.

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Newton’s Law of CoolingW h hi h bWe now have                                                  which can be written as                                       .

Solving for y(t) we obtain the general solution

where we define A = ecwhere we define A = e .

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EXAMPLE

EQUATIONS1.3 Newton’s Law of Cooling for a Cup of 

C ffCoffeeA cup of fast‐food coffee is 180oF when freshly poured. f 2 i i 0 h ff h l dAfter 2 minutes in a room at 70oF, the coffee has cooled 

to 165oF. Find the temperature at any time t and find the i hi h h ff h l d 120 Ftime at which the coffee has cooled to 120oF.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.3 Newton’s Law of Cooling for a Cup of 

C ff

Solution

Coffee

Let y(t) be the temperature of the coffee at time t, we have

Using the initial condition y(0) = 180, gives us

Using the second measured temperature, we have

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.3 Newton’s Law of Cooling for a Cup of 

C ff

Solution

Coffee

We now solve this equation for k.

U i thi l f k lUsing this value for k, we can now solve

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Compound InterestIf a bank agrees to pay you 8% (annual) interest on an investment of $10,000, then at the end of a year, you will have

If a bank agrees to pay you interest twice a year at 8% annually then you will receive (8/2)% interest twice perannually, then you will receive (8/2)% interest twice per year, and at the end of a year you will have

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Compound InterestCompounding interest monthly would pay (8/12)% each month resulting in a balance of

Compounding interest daily would result in a balance of

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Compound InterestIs there is a limit to how much interest can accrue on a given investment at a given interest rate?

If n is the number of times per year that interest is compounded, we wish to calculate the annual percentage yield (APY) under continuous compounding,

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Compound InterestRecall that 

If we make the change of variables n = 0.08m, we have

Continuous compounding would earn approximately 8.3%.

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Continuous compounding would earn approximately 8.3%.

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Compound Interest$Suppose that you invest $P at an annual interest rate r, 

compounded n times per year. Then the value of your investment after t years is

Under continuous compounding (i.e., limit as n→ ∞) this bbecomes

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Compound InterestIf y(t) is the value of your investment after t years, with continuous compounding, the rate of change of y(t) is proportional to y(t). 

F i i i l i f $P hFor an initial investment of $P, we have 

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.5 Depreciation of Assets

(a) Suppose that the value of a $10,000 asset decreases i l f 2 % i dcontinuously at a constant rate of 24% per year. Find 

its worth after 10 years; after 20 years. 

(b) Compare these values to a $10,000 asset that is d i d l i 20 i lidepreciated to no value in 20 years using linear depreciation.

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.5 Depreciation of Assets

SolutionThe value v(t) satisfies v’ = rv, where  r = −0.24. Thus, we havehave

Using the initial value, we haveg

t =10:

t =20:

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t =20:

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7.1 MODELING WITH DIFFERENTIAL EQUATIONS

EXAMPLE

EQUATIONS1.5 Depreciation of Assets

SolutionFor linear depreciation we use the following information.

Using this information we find  

t =10:   v(10) = $5000t 20 (20) $0

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t =20:   v(20) = $0


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