Solutions Manual for
Elementary
Differential
Equations, 10e
William Boyce,
Richard DiPrima (All
Chapters)
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,CHAPTER
1
Introduction
1.1
1.
For y > 3/2, the slopes are negative, therefore the solutions are decreasing. For
y < 3/2, the slopes are positive, hence the solutions are increasing. The equilibrium
solution appears to be y(t) = 3/2, to which all other solutions converge.
1
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,2 Chapter 1. Introduction
3.
For y > −3/2, the slopes are positive, therefore the solutions increase. For y <
−3/2, the slopes are negative, and hence the solutions decrease. All solutions
appear to diverge away from the equilibrium solution y(t) = −3/2.
5.
For y > −1/2, the slopes are positive, and hence the solutions increase. For y <
−1/2, the slopes are negative, and hence the solutions decrease. All solutions
diverge away from the equilibrium solution y(t) = −1/2.
6.
For y > −2, the slopes are positive, and hence the solutions increase. For y < −2,
the slopes are negative, and hence the solutions decrease. All solutions diverge
away from the equilibrium solution y(t) = −2.
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, 1.1 3
8. For all solutions to approach the equilibrium solution y(t) = 2/3, we must have
y 0 < 0 for y > 2/3, and y 0 > 0 for y < 2/3. The required rates are satisfied by the
differential equation y 0 = 2 − 3y.
10. For solutions other than y(t) = 1/3 to diverge from y = 1/3, we must have
y 0 < 0 for y < 1/3, and y 0 > 0 for y > 1/3. The required rates are satisfied by the
differential equation y 0 = 3y − 1.
12.
Note that y 0 = 0 for y = 0 and y = 5. The two equilibrium solutions are y(t) = 0
and y(t) = 5. Based on the direction field, y 0 > 0 for y > 5; thus solutions with
initial values greater than 5 diverge from the solution y(t) = 5. For 0 < y < 5, the
slopes are negative, and hence solutions with initial values between 0 and 5 all
decrease toward the solution y(t) = 0. For y < 0, the slopes are all positive; thus
solutions with initial values less than 0 approach the solution y(t) = 0.
14.
Observe that y 0 = 0 for y = 0 and y = 2. The two equilibrium solutions are y(t) = 0
and y(t) = 2. Based on the direction field, y 0 > 0 for y > 2; thus solutions with
initial values greater than 2 diverge from y(t) = 2. For 0 < y < 2, the slopes are
also positive, and hence solutions with initial values between 0 and 2 all increase
toward the solution y(t) = 2. For y < 0, the slopes are all negative; thus solutions
with initial values less than 0 diverge from the solution y(t) = 0.
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Elementary
Differential
Equations, 10e
William Boyce,
Richard DiPrima (All
Chapters)
Downloaded by: tutorsection | Want to earn $1.236
Distribution of this document is illegal extra per year?
,CHAPTER
1
Introduction
1.1
1.
For y > 3/2, the slopes are negative, therefore the solutions are decreasing. For
y < 3/2, the slopes are positive, hence the solutions are increasing. The equilibrium
solution appears to be y(t) = 3/2, to which all other solutions converge.
1
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Distribution of this document is illegal extra per year?
,2 Chapter 1. Introduction
3.
For y > −3/2, the slopes are positive, therefore the solutions increase. For y <
−3/2, the slopes are negative, and hence the solutions decrease. All solutions
appear to diverge away from the equilibrium solution y(t) = −3/2.
5.
For y > −1/2, the slopes are positive, and hence the solutions increase. For y <
−1/2, the slopes are negative, and hence the solutions decrease. All solutions
diverge away from the equilibrium solution y(t) = −1/2.
6.
For y > −2, the slopes are positive, and hence the solutions increase. For y < −2,
the slopes are negative, and hence the solutions decrease. All solutions diverge
away from the equilibrium solution y(t) = −2.
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, 1.1 3
8. For all solutions to approach the equilibrium solution y(t) = 2/3, we must have
y 0 < 0 for y > 2/3, and y 0 > 0 for y < 2/3. The required rates are satisfied by the
differential equation y 0 = 2 − 3y.
10. For solutions other than y(t) = 1/3 to diverge from y = 1/3, we must have
y 0 < 0 for y < 1/3, and y 0 > 0 for y > 1/3. The required rates are satisfied by the
differential equation y 0 = 3y − 1.
12.
Note that y 0 = 0 for y = 0 and y = 5. The two equilibrium solutions are y(t) = 0
and y(t) = 5. Based on the direction field, y 0 > 0 for y > 5; thus solutions with
initial values greater than 5 diverge from the solution y(t) = 5. For 0 < y < 5, the
slopes are negative, and hence solutions with initial values between 0 and 5 all
decrease toward the solution y(t) = 0. For y < 0, the slopes are all positive; thus
solutions with initial values less than 0 approach the solution y(t) = 0.
14.
Observe that y 0 = 0 for y = 0 and y = 2. The two equilibrium solutions are y(t) = 0
and y(t) = 2. Based on the direction field, y 0 > 0 for y > 2; thus solutions with
initial values greater than 2 diverge from y(t) = 2. For 0 < y < 2, the slopes are
also positive, and hence solutions with initial values between 0 and 2 all increase
toward the solution y(t) = 2. For y < 0, the slopes are all negative; thus solutions
with initial values less than 0 diverge from the solution y(t) = 0.
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