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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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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.
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.
7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONS
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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
Compound InterestIf a bank agrees to pay you 8% (annual) interest on an investment of $10,000, then at the end of a year, you will have
If a bank agrees to pay you interest twice a year at 8% annually then you will receive (8/2)% interest twice perannually, then you will receive (8/2)% interest twice per year, and at the end of a year you will have
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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
Compound InterestCompounding interest monthly would pay (8/12)% each month resulting in a balance of
Compounding interest daily would result in a balance of
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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
Compound InterestIs there is a limit to how much interest can accrue on a given investment at a given interest rate?
If n is the number of times per year that interest is compounded, we wish to calculate the annual percentage yield (APY) under continuous compounding,
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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
Compound InterestRecall that
If we make the change of variables n = 0.08m, we have
Continuous compounding would earn approximately 8.3%.
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Continuous compounding would earn approximately 8.3%.
7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
Compound Interest$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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7.1 MODELING WITH DIFFERENTIAL EQUATIONSEQUATIONS
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:
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