SOLUTION MANUAL FOR
DIFFERENTIAL EQUATIONS AND
LINEAR ALGEBRA, 4TH EDITION
STEPHEN GOODE, SCOTT ANNIN
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Chapter 1 Solutions
Solutions to Section 1.1
True-False Review:
(a): FALSE. A derivative must involve some derivative of the function y = f (x), not necessarily the first
derivative.
(b): FALSE. The order of a differential equation is the order of the highest, not the lowest, derivative
appearing in the differential equation.
(c): FALSE. This differential equation has order two, since the highest order derivative that appears in the
equation is the second order expression y .
(d): FALSE. The carrying capacity refers to the maximum population size that the environment can
support in the long run; it is not related to the initial population in any way.
(e): TRUE. The value y(0) is called an initial condition to the differential equation for y(t).
(f ): TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference
between the object’s temperature and the medium’s temperature. Since that difference is greater for the
object at 100◦ F than the object at 90◦ F , the object whose temperature is 100◦ F has a greater rate of
cooling.
(g): FALSE. The temperature of the object is given by T (t) = Tm + ce−kt , where Tm is the temperature
of the medium, and c and k are constants. Since e−kt = 0, we see that T (t) = Tm for all times t. The
temperature of the object approaches the temperature of the surrounding medium, but never equals it.
(h): TRUE. Since the temperature of the coffee is falling, the temperature difference between the coffee
and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee
has already decreased.
(i): FALSE. The slopes of the two curves are negative reciprocals of each other.
(j): TRUE. If the original family of parallel lines have slopes k for k = 0, then the family of orthogonal tra-
jectories are parallel lines with slope − k1 . If the original family of parallel lines are vertical (resp. horizontal),
then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.
(k): FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family
of lines passing through the origin.
(l): TRUE. If v(t) denotes the velocity of the object at time t and a(t) denotes the velocity of the object
at time t, then we have a(t) = v (t), which is a differential equation for the unknown function v(t).
(m): FALSE. The restoring force is directed in the direction opposite to the displacement from the equi-
librium position.
(n): TRUE. The allometric relationship B = B0 m3/4 , where B0 is a constant, relates the metabolic rate
and total body mass for any species.
Problems:
1. The order is 2.
2. The order is 1.
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3. The order is 3.
4. The order is 2.
5. We compute the first three derivatives of y(t) = ln t:
dy 1 d2 y 1 d3 y 2
= , = − 2, = 3.
dt t dt2 t dt3 t
Therefore,
3
dy 2 d3 y
2 = 3
= 3,
dt t dt
as required.
6. We compute the first two derivatives of y(x) = x/(x + 1):
dy 1 d2 y 2
= and =− .
dx (x + 1)2 dx 2 (x + 1)3
Then
d2 y x 2 x3 + 2x2 + x − 2 (x + 1) + (x3 + 2x2 − 3) 1 x3 + 2x2 − 3
y+ 2
= − 3
= 3
= 3
= 2
+ ,
dx x + 1 (x + 1) (x + 1) (x + 1) (x + 1) (1 + x)3
as required.
7. We compute the first two derivatives of y(x) = ex sin x:
dy d2 y
= ex (sin x + cos x) and = 2ex cos x.
dx dx2
Then
d2 y
2y cot x − = 2(ex sin x) cot x − 2ex cos x = 0,
dx2
as required.
dT d
8. (T − Tm )−1 = −k =⇒ (ln |T − Tm |) = −k. The preceding equation can be integrated directly to
dt dt
yield ln |T − Tm | = −kt + c1 . Exponentiating both sides of this equation gives |T − Tm | = e−kt+c1 , which
can be written as
T − Tm = ce−kt ,
where c = ±ec1 . Rearranging yields T (t) = Tm + ce−kt .
9. After 4 p.m. In the first two hours after noon, the water temperature increased from 50◦ F to 55◦
F, an increase of five degrees. Because the temperature of the water has grown closer to the ambient air
temperature, the temperature difference |T − Tm | is smaller, and thus, the rate of change of the temperature
of the water grows smaller, according to Newton’s Law of Cooling. Thus, it will take longer for the water
temperature to increase another five degrees. Therefore, the water temperature will reach 60◦ F more than
two hours later than 2 p.m., or after 4 p.m.
10. The object temperature cools a total of 40◦ F during the 40 minutes, but according to Newton’s Law of
Cooling, it cools faster in the beginning (since |T − Tm | is greater at first). Thus, the object cooled half-way
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from 70◦ F to 30◦ F in less than half the total cooling time. Therefore, it took less than 20 minutes for the
object to reach 50◦ F.
dy dy x
11. The given family of curves satisfies: x2 + 9y 2 = c =⇒ 2x + 18y = 0 =⇒ =− .
dx dx 9y
Orthogonal trajectories satisfy:
dy 9y 1 dy 9 d 9
= =⇒ = =⇒ (ln |y|) = =⇒ ln |y| = 9 ln |x| + c1 =⇒ y = kx9 , where k = ±ec1
dx x y dx x dx x
.
y(x)
0.8
0.4
x
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.4
-0.8
Figure 0.0.1: Figure for Problem 11
y
12. Given family of curves satisfies: y = cx2 =⇒ c = . Hence,
x2
dy y 2y
= 2cx = c 2 x = .
dx x x
Orthogonal trajectories satisfy:
dy x dy d 2 1
=− =⇒ 2y = −x =⇒ (y ) = −x =⇒ y 2 = − x2 + c1 =⇒ 2y 2 + x2 = c2 ,
dx 2y dx dx 2
where c2 = 2c1 .
c dy dy y
13. Given a family of curves satisfies: y = =⇒ x + y = 0 =⇒ =− .
x dx dx x
Orthogonal trajectories satisfy:
dy x dy d 1 2 1 1
= =⇒ y = x =⇒ y = x =⇒ y 2 = x2 + c1 =⇒ y 2 − x2 = c2 , where c2 = 2c1 .
dx y dx dx 2 2 2
y
14. The given family of curves satisfies: y = cx5 =⇒ c = . Hence,
x5
dy y 5y
= 5cx4 = 5 5 x4 = .
dx x x
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