Section 9.1 Modeling with Differential Equations
19. (a) What can you say about a solution of the equation y0= −y2 just by looking at the differential equation?
(b) Verify that all members of the family y = 1/(x + C) are solutions of the equation in part (a).
(c) Can you think of a solution of the differential equation y0= −y2 that is not a member of the family in part (b)?
(d) Find a solution of the initial-value problem
y0 = −y2 y(0) = 0.5 Solution:
804 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
5. (a) = sin ⇒ 0= cos ⇒ 00= − sin .
LHS = 00+ = − sin + sin = 0 6= sin , so = sin is not a solution of the differential equation.
(b) = cos ⇒ 0= − sin ⇒ 00= − cos .
LHS = 00+ = − cos + cos = 0 6= sin , so = cos is not a solution of the differential equation.
(c) = 12 sin ⇒ 0 = 12( cos + sin ) ⇒ 00= 12(− sin + cos + cos ).
LHS = 00+ = 12(− sin + 2 cos ) +12 sin = cos 6= sin , so = 12 sin is not a solution of the differential equation.
(d) = −12 cos ⇒ 0 = −12(− sin + cos ) ⇒ 00= −12(− cos − sin − sin ).
LHS = 00+ = −12(− cos − 2 sin ) +
−12 cos
= sin = RHS, so = −12 cos is a solution of the differential equation.
6. (a) = ln +
⇒ 0= · (1) − (ln + )
2 =1 − ln −
2 .
LHS = 20+ = 2·1 − ln −
2 + ·ln +
= 1 − ln − + ln + = 1 = RHS, so is a solution of the differential equation.
(b) A few notes about the graph of = (ln + ):
(1) There is a vertical asymptote of = 0.
(2) There is a horizontal asymptote of = 0.
(3) = 0 ⇒ ln + = 0 ⇒ = −, so there is an -intercept at −.
(4) 0= 0 ⇒ ln = 1 − ⇒ = 1−, so there is a local maximum at = 1−.
(c) (1) = 2 ⇒ 2 =ln 1 +
1 ⇒ 2 = , so the solution is = ln + 2
[shown in part (b)].
(d) (2) = 1 ⇒ 1 =ln 2 +
2 ⇒ 2 + ln 2 + ⇒ = 2 − ln 2, so the solution is =ln + 2 − ln 2
[shown in part (b)].
7. (a) Since the derivative 0= −2is always negative (or 0 if = 0), the function must be decreasing (or equal to 0) on any interval on which it is defined.
(b) = 1
+ ⇒ 0= − 1
( + )2. LHS = 0= − 1
( + )2 = −
1
+
2
= −2= RHS
(c) = 0 is a solution of 0= −2that is not a member of the family in part (b).
(d) If () = 1
+ , then (0) = 1 0 + = 1
. Since (0) = 05, 1
= 1
2 ⇒ = 2, so = 1
+ 2.
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21. A population is modeled by the differential equation dP
dt = 1.2P
1 − P
4200
(a) For what values of P is the population increasing?
(b) For what values of P is the population decreasing?
(c) What are the equilibrium solutions?
Solution:
SECTION 9.1 MODELING WITH DIFFERENTIAL EQUATIONS ¤ 805
8. (a) If is close to 0, then 3is close to 0, and hence, 0is close to 0. Thus, the graph of must have a tangent line that is nearly horizontal. If is large, then 3is large, and the graph of must have a tangent line that is nearly vertical.
(In both cases, we assume reasonable values for .)
(b) = ( − 2)−12 ⇒ 0= ( − 2)−32. RHS = 3= [
− 2−12]3= ( − 2)−32= 0= LHS
(c) When is close to 0, 0is also close to 0.
As gets larger, so does |0|.
(d) (0) = ( − 0)−12= 1√
and (0) = 2 ⇒ √
= 12 ⇒ = 14, so =1
4− 2−12. 9. (a)
= 12
1 −
4200
. Now
0 ⇒ 1 −
4200 0 [assuming that 0] ⇒
4200 1 ⇒
4200 ⇒ the population is increasing for 0 4200.
(b)
0 ⇒ 4200 (c)
= 0 ⇒ = 4200 or = 0 10. (a) = ⇒ 0= 0, so
= 4− 63+ 52 ⇔ 0 = 4− 63+ 52 ⇔ 2
2− 6 + 5
= 0 ⇔
2( − 1)( − 5) = 0 ⇔ = 0, 1, or 5 (b) is increasing ⇔
0 ⇔ 2( − 1)( − 5) 0 ⇔ ∈ (−∞ 0) ∪ (0 1) ∪ (5 ∞) (c) is decreasing ⇔
0 ⇔ ∈ (1 5)
11. (a) This function is increasing and also decreasing. But = ( − 1)2≥ 0 for all , implying that the graph of the solution of the differential equation cannot be decreasing on any interval.
(b) When = 1, = 0, but the graph does not have a horizontal tangent line.
12. The graph for this exercise is shown in the figure at the right.
A. 0= 1 + 1for points in the first quadrant, but we can see that 0 0for some points in the first quadrant.
B. 0= −2 = 0 when = 0, but we can see that 0 0for = 0.
Thus, equations A and B are incorrect, so the correct equation is C.
C. 0= 1 − 2 seems reasonable since:
(1) When = 0, 0could be 1.
(2) When 0, 0could be greater than 1.
(3) Solving 0= 1 − 2 for gives us = 1 − 0
2 . If 0takes on small negative values, then as → ∞, → 0+, as shown in the figure.
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25. Match the differential equations with the solution graphs labeled IIV. Give reasons for your choices.
(a) y0= 1 + x2+ y2 (b) y0 = xe−x2−y2 (c) y0 = 1
1+ex2 +y2 (d) y0 = sin(xy) cos(xy)
SeCtion9.2 Direction Fields and Euler’s Method 591
Unfortunately, it’s impossible to solve most differential equations in the sense of obtain- ing an explicit formula for the solution. In this section we show that, despite the absence of an explicit solution, we can still learn a lot about the solution through a graphical approach (direction fields) or a numerical approach (Euler’s method).
expresses Newton’s Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling?
(c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).
15. Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function Pstd, the performance of someone learning a skill as a function of the training time t. The derivative dPydt represents the rate at which performance improves.
(a) When do you think P increases most rapidly? What hap- pens to dPydt as t increases? Explain.
(b) If M is the maximum level of performance of which the learner is capable, explain why the differential equation
dP
dt − ksM 2 Pd k a positive constant is a reasonable model for learning.
(c) Make a rough sketch of a possible solution of this dif- ferential equation.
16. Von Bertalanffy’s equation states that the rate of growth in length of an individual fish is proportional to the difference between the current length L and the asymptotic length L` (in centimeters).
(a) Write a differential equation that expresses this idea.
(b) Make a rough sketch of the graph of a solution of a typi- cal initial-value problem for this differential equation.
17. Differential equations have been used extensively in the study of drug dissolution for patients given oral medications. One such equation is the Weibull equation for the concentration cstd of the drug:
dc dt − k
tb scs2cd
where k and cs are positive constants and 0 , b , 1. Verify that
cstd − cs
(
1 2 e2 t12b)
is a solution of the Weibull equation for t . 0, where
− kys1 2 bd. What does the differential equation say about how drug dissolution occurs?
12. The function with the given graph is a solution of one of the following differential equations. Decide which is the correct equation and justify your answer.
0 x
y
a. y9 − 1 1 xy B. y9 − 22xy C. y9 − 1 2 2xy 13. Match the differential equations with the solution graphs
labeled I–IV. Give reasons for your choices.
(a) y9 − 1 1 x21y2 (b) y9 − xe2x22y2
(c) y9 − 1
1 1 ex21y2 (d) y9 − sinsxyd cossxyd
y
x x
y
x y
x y
I II
III IV
0
0
0 0
y
x x
y
x y
x y
I II
III IV
0
0
0 0
14. Suppose you have just poured a cup of freshly brewed coffee with temperature 95°C in a room where the tempera ture is 20°C.
(a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain.
(b) Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that
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1
Solution:
806 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
13. (a) 0= 1 + 2+ 2≥ 1 and 0→ ∞ as → ∞. The only curve satisfying these conditions is labeled III.
(b) 0= −2−2 0if 0 and 0 0if 0. The only curve with negative tangent slopes when 0 and positive tangent slopes when 0 is labeled I.
(c) 0= 1
1 + 2+2 0and 0→ 0 as → ∞. The only curve satisfying these conditions is labeled IV.
(d) 0= sin() cos() = 0if = 0, which is the solution graph labeled II.
14. (a) The coffee cools most quickly as soon as it is removed from the heat source. The rate of cooling decreases toward 0 since the coffee approaches room temperature.
(b)
= ( − ), where is a proportionality constant, is the temperature of the coffee, and is the room temperature. The initial condition is (0) = 95◦C. The answer and the model support each other because as approaches , approaches 0, so the model seems appropriate.
(c)
15. (a) increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As increases, we would expect to remain positive, but decrease. This is because as time progresses, the only points left to learn are the more difficult ones.
(b)
= ( − ) is always positive, so the level of performance is increasing. As gets close to , gets close to 0; that is, the performance levels off, as explained in part (a).
(c)
16. (a)
= (∞− ). Assuming ∞ , we have 0 and
0for all .
(b)
17.If () =
1 − −1−
= − −1−for 0, where 0, 0, 0 1, and = (1 − ), then
=
0 − −1−·
(−1−)
= −−1−· (−)(1 − )−=(1 − )
−1−=
(− ). The equation for indicates that as increases, approaches . The differential equation indicates that as increases, the rate of increase of decreases steadily and approaches 0 as approaches .
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29. Differential equations have been used extensively in the study of drug dissolution for patients given oral medications.
One such equation is the Weibull equation for the concentration c(t) of the drug:
dc dt = k
tb(cs− c) where k and csare positive constants and 0 < b < 1. Verify that
c(t) = cs(1 − e−αt1−b)
is a solution of the Weibull equation for t > 0, where α = k/(1 − b). What does the differential equation say about how drug dissolution occurs?
Solution:
806 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
13. (a) 0= 1 + 2+ 2≥ 1 and 0→ ∞ as → ∞. The only curve satisfying these conditions is labeled III.
(b) 0= −2−2 0if 0 and 0 0if 0. The only curve with negative tangent slopes when 0 and positive tangent slopes when 0 is labeled I.
(c) 0= 1
1 + 2+2 0and 0→ 0 as → ∞. The only curve satisfying these conditions is labeled IV.
(d) 0= sin() cos() = 0if = 0, which is the solution graph labeled II.
14. (a) The coffee cools most quickly as soon as it is removed from the heat source. The rate of cooling decreases toward 0 since the coffee approaches room temperature.
(b)
= ( − ), where is a proportionality constant, is the temperature of the coffee, and is the room temperature. The initial condition is (0) = 95◦C. The answer and the model support each other because as approaches , approaches 0, so the model seems appropriate.
(c)
15. (a) increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As increases, we would expect to remain positive, but decrease. This is because as time progresses, the only points left to learn are the more difficult ones.
(b)
= ( − ) is always positive, so the level of performance is increasing. As gets close to , gets close to 0; that is, the performance levels off, as explained in part (a).
(c)
16. (a)
= (∞− ). Assuming ∞ , we have 0 and
0for all .
(b)
17.If () =
1 − −1−
= − −1−for 0, where 0, 0, 0 1, and = (1 − ), then
=
0 − −1−·
(−1−)
= −−1−· (−)(1 − )−= (1 − )
−1− =
(− ). The equation for indicates that as increases, approaches . The differential equation indicates that as increases, the rate of increase of decreases steadily and approaches 0 as approaches .
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