AP Calculus BC
Review of Ch 6 – Differential Equations
Name:_________________________
12 January 2015
The problems below are a sampling of review questions. See the recap from January 3rd for a more
complete list of topics.
First, two types of problems we have not done in awhile:
1. Integrate ∫ ln(x)dx
2. If f (x) =
x
2
∫ a e t dt (something we can’t integrate with any of our methods) and f(1) = 5, what is
f(3)?
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Several Practice Problems – for problems 3 – 6 you should
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a) Pick a method and solve the differential equation symbolically
b) Graph the slope field
c) Graph the solution from part a on the slope field
d) Find a solution using Euler’s method, and
e) Find the exact value using your solution from part a
dy
4x
=−
3.
at (2, 3), find y(1.7)
dx
y
1 dh
3
4.
with h(–1) = 4, find h(–1.4)
2 dt = cos(t +1)
6t
dy 5y
1
5.
− 2 = 2 which goes through the point (1, 0,4), find y(1.5)
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dx x
x
dP
6.
= .2P(15 − 3P) with P(0) = 2, find P(0.3)
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dt
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A “You have to use Laplace Transforms” Problem
7. y" + 2 y" +10 = 0 with y(0) = 0.1 and y"(0) = 0
Application problems
8. Biomass is a measure of an amount of living matter in an ecosystem. Suppose the biomass s(t)
€ at a rate of about 3.5 tons per year, and decreases by about 5%
in a given ecosystem increases
ds
per year. This situation can be modeled by the differential equation
= 3.5 − .05s
dt
a) Solve the differential equation
b) Use a graphing utility to graph the slope field for the differential equation
c) Explain what happens as t →∞
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9. An automobile gets 28 miles per gallon of gasoline for speeds up to 50 miles per hour. Over 50
miles per hour, the number of miles per gallons drops at a rate of 12 percent for each 10 miles
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per hour.
a) S is the speed and m is the number of miles per gallon. Find m as a function of S by solving
dm
the differential equation
= −0.012m , S > 50.
dS
b) Use the function in part a to complete the table.
Speed
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Miles per Gallon
50
55
60
65
70
Review of Ch 6 – Differential Equations
page 2
Answer Key of Analytic/Symbolic Solutions
3
2
∫1 e t dt
1. x ln(x) – x + C
2. f (3) = 5 +
4. h(t) = 2sin(t3 + 1) + 4
5. y =
7. y = 0.1e–tcos(3t)
8, s = 70 − Ae −.05t
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1
5
(
3. 4x2 + y2 = 25
5
6. P =
1+1.5e −3t
9. s = 51.019e −.012t
)
e −5(1− x ) +1
1
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