**Unformatted text preview: **1
Solutions to Section 1.1
True-False Review:
1. FALSE. A derivative must involve some derivative of the function y = f (x), not necessarily the ﬁrst
derivative.
2. TRUE. The initial conditions accompanying a diﬀerential equation consist of the values of y, y , . . . at
t = 0.
3. TRUE. If we deﬁne positive velocity to be oriented downward, then
dv
= g,
dt
where g is the acceleration due to gravity.
4. TRUE. We can justify this mathematically by starting from a(t) = g , and integrating twice to get
1
v (t) = gt + c, and then s(t) = gt2 + ct + d, which is a quadratic equation.
2
5. FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium
position.
6. 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.
7. 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.
8. 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.
9. FALSE. The slopes of the two curves are negative reciprocals of each other.
10. TRUE. If the original family of parallel lines have slopes k for k = 0, then the family of orthogonal tra1
jectories are parallel lines with slope − k . If the original family of parallel lines are vertical (resp. horizontal),
then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.
11. FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family
of lines passing through the origin.
Problems:
d2 y
1. Starting from the diﬀerential equation 2 = g , where g is the acceleration of gravity and y is the unknown
dt
position function, we integrate twice to obtain the general equations for the velocity and the position of the
object:
dy
gt2
= gt + c1 and y (t) =
+ c1 t + c2 ,
dt
2
where c1 , c2 are constants of integration. Now we impose the initial conditions: y (0) = 0 implies that c2 = 0, 2
and dy
dt (0) = 0 implies that c1 = 0. Hence, the solution to the initial-value problem is
y (t) = gt2
.
2 The object hits the ground at the time t0 for which y (t0 ) = 100. Hence 100 =
s, where we have taken g = 9.8 ms −2 . gt2
0
2, so that t0 = 200
g ≈ 4.52 d2 y
2. Starting from the diﬀerential equation 2 = g , where g is the acceleration of gravity and y is the unknown
dt
position function, we integrate twice to obtain the general equations for the velocity and the position of the
ball, respectively:
dy
1
= gt + c and y (t) = gt2 + ct + d,
dt
2
where c, d are constants of integration. Setting y = 0 to be at the top of the boy’s head (and positive
direction downward), we know that y (0) = 0. Since the object hits the ground 8 seconds later, we have that
y (8) = 5 (since the ground lies at the position y = 5). From the values of y (0) and y (8), we ﬁnd that d = 0
5 − 32g
and 5 = 32g + 8c. Therefore, c =
.
8
(a) The ball reaches its maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore,
t=− c
32g − 5
=
≈ 3.98 s.
g
8g (b) To ﬁnd the maximum height of the tennis ball, we compute
y (3.98) ≈ −253.51 feet.
So the ball is 253.51 feet above the top of the boy’s head, which is 258.51 feet above the ground.
d2 y
3. Starting from the diﬀerential equation 2 = g , where g is the acceleration of gravity and y is the unknown
dt
position function, we integrate twice to obtain the general equations for the velocity and the position of the
rocket, respectively:
dy
1
= gt + c and y (t) = gt2 + ct + d,
dt
2
where c, d are constants of integration. Setting y = 0 to be at ground level, we know that y (0) = 0. Thus,
d = 0.
(a) The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the
c
time that the rocket achieves its maximum height is t = − . At this time, y (t) = −90 (the negative sign
g
accounts for the fact that the positive direction is chosen to be downward). Hence,
2
c
1
c
c
c2
c2
c2
−90 = y −
=g−
+c −
=
−
=− .
g
2
g
g
2g
g
2g
√
Solving this for c, we ﬁnd that c = ± 180g . However, since c represents the initial velocity of the rocket,
and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose
√
c = − 180g ≈ −42.02 ms−1 , and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 42.02 ms−1 . 3
(b) The time that the rocket reaches its maximum height is t = − c
−42.02
≈−
= 4.28 s.
g
9.81 d2 y
4. Starting from the diﬀerential equation 2 = g , where g is the acceleration of gravity and y is the unknown
dt
position function, we integrate twice to obtain the general equations for the velocity and the position of the
rocket, respectively:
dy
1
= gt + c and y (t) = gt2 + ct + d,
dt
2
where c, d are constants of integration. Setting y = 0 to be at the level of the platform (with positive
direction downward), we know that y (0) = 0. Thus, d = 0.
(a) The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the
c
time that the rocket achieves its maximum height is t = − . At this time, y (t) = −85 (this is 85 m above
g
the platform, or 90 m above the ground). Hence,
2
c
1
c
c
c2
c2
c2
−85 = y −
=g−
+c −
=
−
=− .
g
2
g
g
2g
g
2g √
Solving this for c, we ﬁnd that c = ± 170g . However, since c represents the initial velocity of the rocket,
and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose
√
c = − 170g ≈ −40.84 ms−1 , and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 40.84 ms−1 .
c
−40.84
(b) The time that the rocket reaches its maximum height is t = − ≈ −
= 4.16 s.
g
9.81
5. If y (t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
d2 y
dy
= g, y (0) = 0,
(0) = −2,
dt2
dt
where g is the acceleration of gravity and y is the unknown position function. We integrate this diﬀerential
equation twice to obtain the general equations for the velocity and the position of the object:
dy
= gt + c1
dt and y (t) = gt2
+ c1 t + c2 .
2 Now we impose the initial conditions: since y (0) = 0, we have c2 = 0. Moreover, since
c1 = −2. Hence the solution to the initial-value problem is y (t) =
Consequently, h = dy
(0) = −2, we have
dt gt2
− 2t. We are given that y (10) = h.
2 g (10)2
− 2 · 10 =⇒ h = 10(5g − 2) ≈ 470 m where we have taken g = 9.8 ms−2 .
2 6. If y (t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
d2 y
= g,
dt2 y (0) = 0, dy
(0) = v0 .
dt 4
We integrate the diﬀerential equation twice to obtain the velocity and position functions, respectively:
dy
= gt + c1
dt and y (t) = gt2
+ c1 t + c2 .
2 Now we impose the initial conditions. Since y (0) = 0, we have c2 = 0. Moreover, since
c1 = v0 . Hence the solution to the initial-value problem is y (t) =
Consequently, h = gt2 + v0 t0 . Solving for v0 yields
0
v0 = dy
(0) = v0 , we have
dt gt2
+ v0 t. We are given that y (t0 ) = h.
2 2h − gt2
0
.
2t0 7. From y (t) = A cos (ω t − φ), we obtain
dy
= −Aω sin (ω t − φ)
dt
Hence, and d2 y
= −Aω 2 cos (ω t − φ).
dt2 d2 y
+ ω 2 y = −Aω 2 cos (ω t − φ) + Aω 2 cos (ω t − φ) = 0.
dt2 Substituting y (0) = a, we obtain a = A cos(−φ) = A cos(φ). Also, from dy (0) = 0, we obtain 0 =
dt
−Aω sin(−φ) = Aω sin(φ). Since A = 0 and ω = 0 and |φ| < π , we have φ = 0. It follows that a = A.
8. Taking derivatives of y (t) = c1 cos (ω t) + c2 sin (ω t), we obtain
dy
= −c1 ω sin (ω t) + c2 ω cos (ω t)
dt
and d2 y
= −c1 ω 2 cos (ω t) − c2 ω 2 sin (ω t) = −ω 2 [c1 cos (ω t) + c2 cos (ω t)] = −ω 2 y.
dt2 d2 y
Consequently,
+ ω 2 y = 0. To determine the amplitude of the motion we write the solution to the
dt2
diﬀerential equation in the equivalent form:
c2
2 + c2 c1
y (t) = c1
cos (ω t) + 2
sin (ω t) .
2
c2 + c2
c1 + c2
1
2
2
We can now deﬁne an angle φ by cos φ = c1
c2 + c2
1
2 and c2 sin φ = c2 + c2
1
2 . Then the expression for the solution to the diﬀerential equation is
y (t) = c2 + c2 [cos (ω t) cos φ + sin (ω t) sin φ] = c2 + c2 cos (ω t + φ).
1
2
1
2
Consequently the motion corresponds to an oscillation with amplitude A =
c2 + c2 .
1
2 5
9. We compute the ﬁrst three derivatives of y (t) = ln t:
dy
1
=,
dt
t d2 y
1
= − 2,
2
dt
t Therefore,
2
as required. dy
dt 3 = d3 y
2
= 3.
3
dt
t 2
d3 y
= 3,
t3
dt 10. We compute the ﬁrst two derivatives of y (x) = x/(x + 1):
dy
1
=
dx
(x + 1)2 d2 y
2
=−
.
dx2
(x + 1)3 and Then
y+ d2 y
x
2
x3 + 2x2 + x − 2
(x + 1) + (x3 + 2x2 − 3)
1
x3 + 2x2 − 3
=
−
=
=
=
+
,
dx2
x + 1 (x + 1)3
(x + 1)3
(x + 1)3
(x + 1)2
(1 + x)3 as required.
11. We compute the ﬁrst two derivatives of y (x) = ex sin x:
dy
= ex (sin x + cos x)
dx
Then
2y cot x − and d2 y
= 2ex cos x.
dx2 d2 y
= 2(ex sin x) cot x − 2ex cos x = 0,
dx2 as required.
dT
d
= −k , we obtain
(ln |T − Tm |) = −k . The preceding equation can
dt
dt
be integrated directly to yield ln |T − Tm | = −kt + c1 . Exponentiating both sides of this equation gives
|T − Tm | = e−kt+c1 , which can be written as
12. Starting from (T − Tm )−1 T − Tm = ce−kt ,
where c = ±ec1 . Rearranging this, we conclude that T (t) = Tm + ce−kt .
13. 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.
14. 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
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. 6
15. Applying implicit diﬀerentiation to the given family of curves x2 + 4y 2 = c with respect to x gives
2x + 8y dy
= 0.
dx Therefore,
dy
x
=− .
dx
4y
Therefore, the collection of orthogonal trajectories satisﬁes:
dy
4y
=
dx
x =⇒ 1 dy
4
=
y dx
x =⇒ d
4
(ln |y |) =
dx
x =⇒ ln |y | = 4 ln |x| + c1 =⇒ y = kx4 , where k = ±ec1 .
y(x) 0.8
0.4
x -1.5 -0.5 -1.0 1.0 0.5 1.5 -0.4
-0.8 Figure 1: Figure for Problem 15
16. Diﬀerentiation of the given family of curves y = c
with respect to x gives
x dy
c
1c
y
=− 2 =− · =− .
dx
x
xx
x
Therefore, the collection of orthogonal trajectories satisﬁes:
dy
x
dy
d 12
=
=⇒ y
= x =⇒
y = x =⇒
dx
y
dx
dx 2 12
1
y = x2 + c1
2
2 where c2 = 2c1 .
y(x) 4 2 x
-4 -2 2 4 -2 -4 Figure 2: Figure for Problem 16 =⇒ y 2 − x2 = c2 , 7
y
. Hence, diﬀerentiation leads to
x2
y
dy
2y
= 2cx = 2 2 x =
.
dx
x
x 17. Solving the equation y = cx2 for c gives c = Therefore, the collection of orthogonal trajectories satisﬁes:
dy
x
=−
dx
2y =⇒ 2y dy
= −x
dx d2
(y ) = −x
dx =⇒ 1
y 2 = − x2 + c1
2 =⇒ =⇒ 2y 2 + x2 = c2 , where c2 = 2c1 .
y(x)
2.0
1.6
1.2
0.8
0.4
x
-1 -2 1 2 -0.4
-0.8
-1.2
-1.6
-2.0 Figure 3: Figure for Problem 17
y
. Hence,
x4
y
dy
4y
= 4cx3 = 4 4 x3 =
.
dx
x
x 18. Solving the equation y = cx4 for c gives c = Therefore, the collection of orthogonal trajectories satisﬁes:
dy
x
=−
dx
4y =⇒ 4y dy
= −x
dx =⇒ d
(2y 2 ) = −x
dx =⇒ 1
2y 2 = − x2 + c1
2 =⇒ 4y 2 + x2 = c2 , where c2 = 2c1 .
y(x) 0.8
0.4
x
-1.5 1.0 -1.0 1.5 -0.4
-0.8 Figure 4: Figure for Problem 18
dy
19. Implicit diﬀerentiation of the given family of curves y 2 = 2x + c with respect to x gives 2y dx = 2. That
dy
1
is,
= . Therefore, the collection of orthogonal trajectories satisﬁes:
dx
y dy
= −y
dx =⇒ y −1 dy
= −1
dx =⇒ d
(ln |y |) = −1
dx =⇒ ln |y | = −x + c1 =⇒ y = ke−x , 8
y(x) 4 3 2 1 x -1 1 2 4 3 -1 -2 -3 -4 Figure 5: Figure for Problem 19
where k = ±ec1 . dy
20. Diﬀerentiating the given family of curves y = cex with respect to x gives
= cex = y . Therefore, the
dx
collection of orthogonal trajectories satisﬁes:
dy
1
dy
d 12
12
=−
=⇒ y
= −1 =⇒
y = −1 =⇒
y = −x + c1 =⇒ y 2 = −2x + c2 ,
dx
y
dx
dx 2
2
where c2 = 2c1 .
y(x) 2 1 x
1 -1 -1 -2 Figure 6: Figure for Problem 20
21. Diﬀerentiating the given family of curves y = mx + c with respect to x gives
collection of orthogonal trajectories satisﬁes:
dy
1
=−
dx
m =⇒ y=− dy
dx = m. Therefore, the 1
x + c1 .
m dy
y
22. Diﬀerentiating the given family of curves y = cxm with respect to x gives
= cmxm−1 . Since c = m ,
dx
x
dy
my
we have
=
. Therefore, the collection of orthogonal trajectories satisﬁes:
dx
x
dy
x
dy
x
d 12
x
12
12
1
=−
=⇒ y
=−
=⇒
y =−
=⇒
y =−
x +c1 =⇒ y 2 = − x2 +c2 ,
dx
my
dx
m
dx 2
m
2
2m
m 9
where c2 = 2c1 . dy
23. Implicit diﬀerentiation of the given family of curves y 2 + mx2 = c with respect to x gives 2y +2mx = 0.
dx
dy
mx
That is,
=−
. Therefore, the collection of orthogonal trajectories satisﬁes:
dx
y
dy
y
=
dx
mx =⇒ y −1 dy
1
=
dx
mx d
1
(ln |y |) =
dx
mx =⇒ =⇒ m ln |y | = ln |x| + c1 =⇒ y m = c2 x, where c2 = ±ec1 .
dy
24. Implicit diﬀerentiation of the given family of curves y 2 = mx + c with respect to x gives 2y
= m.
dx
dy
m
That is,
=
. Therefore, the collection of orthogonal trajectories satisﬁes:
dx
2y
dy
2y
=−
dx
m =⇒ y −1 dy
2
=−
dx
m =⇒ d
2
(ln |y |) = −
dx
m =⇒ ln |y | = − 2
x+c1
m =⇒ 2x y = c2 e− m , where c2 = ±ec1 .
25. Consider the coordinate curve u = x2 + 2y 2 (i.e. u is constant). Diﬀerentiating implicitly with respect to
dy
dy
x
x gives 0 = 2x + 4y . Therefore,
= − . Therefore, the collection of orthogonal trajectories satisﬁes:
dx
dx
2y
dy
2y
=
dx
x =⇒ y −1 dy
2
=
dx
x d
2
(ln |y |) =
dx
x =⇒ =⇒ ln |y | = 2 ln |x| + c1 where c2 = ±ec1 .
y(x) 2.0
1.6
1.2
0.8
0.4
x -2 1 -1 2 -0.4
-0.8
-1.2
-1.6
-2.0 Figure 7: Figure for Problem 25 26. We have m1 = tan (a1 ) = tan (a2 − a) = tan (a2 ) − tan (a)
m2 − tan (a)
=
.
1 + tan (a2 ) tan (a)
1 + m2 tan (a) =⇒ y = c2 x2 , 10
Solutions to Section 1.2
True-False Review:
1. FALSE. The order of a diﬀerential equation is the order of the highest derivative appearing in the
diﬀerential equation.
2. TRUE. This is condition 1 in Deﬁnition 1.2.11.
3. TRUE. This is the content of Theorem 1.2.15.
4. FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c2 cos x, such as
y (x) = sin x. Therefore, c1 cos x + 5c2 cos x does not meet the second requirement set forth in Deﬁnition
1.2.11 for the general solution.
5. FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c1 sin x, such as
y (x) = cos x + sin x. Therefore, c1 cos x + 5c1 sin x does not meet the second requirement set form in
Deﬁnition 1.2.11 for the general solution.
6. TRUE. Since the right-hand side of the diﬀerential equation is a function of x only, we can integrate
both sides n times to obtain the formula for the solution y (x).
Problems:
1. 2, nonlinear.
2. 3, linear.
3. 2, nonlinear.
4. 2, nonlinear.
5. 4, linear.
6. 3, nonlinear.
7. We can quickly compute the ﬁrst two derivatives of y (x):
y (x) = (c1 +2c2 )ex cos 2x +(−2c1 + c2 )ex sin 2x
Then we have and y (x) = (−3c1 +4c2 )ex cos 2x +(−4c1 − 3c2 )ex sin x. y − 2y + 5y = [(−3c1 + 4c2 )ex cos 2x + (−4c1 − 3c2 )ex sin x]−2 [(c1 + 2c2 )ex cos 2x + (−2c1 + c2 )ex sin 2x]+5(c1 ex cos 2x+c2 ex sin 2x),
which cancels to 0, as required. This solution is valid for all x ∈ R.
8. We can quickly compute the ﬁrst two derivatives of y (x):
y (x) = c1 ex − 2c2 e−2x and y (x) = c1 ex + 4c2 e−2x . Then we have
y + y − 2y = (c1 ex + 4c2 e−2x ) + (c1 ex − 2c2 e−2x ) − 2(c1 ex + c2 e−2x ) = 0.
Thus y (x) = c1 ex + c2 e−2x is a solution of the given diﬀerential equation for all x ∈ R. 11
1
1
1
is y (x) = −
= −y 2 . Thus y (x) =
is a solution of the given
x+4
(x + 4)2
x+4
diﬀerential equation for x ∈ (−∞, −4) or x ∈ (−4, ∞).
9. The derivative of y (x) = √
10. The derivative of y (x) = c1 x is y (x) =
diﬀerential equation for all x > 0. √
c1
y
√=
. Thus y (x) = c1 x is a solution of the given
2x
2x 11. We can quickly compute the ﬁrst two derivatives of y (x) = c1 e−x sin (2x):
y (x) = 2c1 e−x cos (2x) − c1 e−x sin (2x) and y (x) = −3c1 e−x sin (2x) − 4c1 e−x cos (2x). Therefore, we have
y + 2y + 5y = −3c1 e−x sin (2x) − 4c1 e−x cos (2x) + 2[2c1 e−x cos (2x) − c1 e−x sin (2x)] + 5[c1 e−x sin (2x)] = 0,
which shows that y (x) = c1 e−x sin (2x) is a solution to the given diﬀerential equation for all x ∈ R.
12. We can quickly compute the ﬁrst two derivatives of y (x) = c1 cosh (3x) + c2 sinh (3x):
y (x) = 3c1 sinh (3x) + 3c2 cosh (3x) and y (x) = 9c1 cosh (3x) + 9c2 sinh (3x). Therefore, we have
y − 9y = [9c1 cosh (3x) + 9c2 sinh (3x)] − 9[c1 cosh (3x) + c2 sinh (3x)] = 0,
which shows that y (x) = c1 cosh (3x) + c2 sinh (3x) is a solution to the given diﬀerential equation for all
x ∈ R.
13. We can quickly compute the ﬁrst two derivatives of y (x) =
y (x) = − 3c1
c2
−2
x4
x and c1
c2
+:
x3
x y (x) = 12c1
2c2
+ 3.
x5
x Therefore, we have
x2 y + 5xy + 3y = x2
which shows that y (x) =
x ∈ (0, ∞). 12c1
2c2
+3
5
x
x
c
3c1
c2
c2
1
+ 5x − 4 − 2 + 3 3 +
= 0,
x
x
x
x c1
c2
+
is a solution to the given diﬀerential equation for all x ∈ (−∞, 0) or
3
x
x √
14. We can quickly compute the ﬁrst two derivatives of y (x) = c1 x + 3x2 :
c1
y (x) = √ + 6x
2x and c1
y (x) = − √ + 6.
4 x3 Therefore, we have
√
c1
c1
√ + 6x + (c1 x + 3x2 ) = 9x2 ,
2x2 y − xy + y = 2x2 − √ + 6 − x
3
2x
4x
√
which shows that y (x) = c1 x + 3x2 is a solution to the given diﬀerential equation for all x > 0. 12
15. We can quickly compute the ﬁrst two derivatives of y (x) = c1 x2 + c2 x3 − x2 sin x:
y (x) = 2c1 x + 3c2 x2 − x2 cos x − 2x sin x and y (x) = 2c1 + 6c2 x + x2 sin x − 2x cos x − 2x cos −2 sin x. Substituting these results into the given diﬀerential equation yields
x2 y − 4xy + 6y = x2 (2c1 + 6c2 x + x2 sin x − 4x cos x − 2 sin x) − 4x(2c1 x + 3c2 x2 − x2 cos x − 2x sin x) + 6(c1 x2 + c2 x3 − x2 sin x) = 2c1 x2 + 6c2 x3 + x4 sin x − 4x3 cos x − 2x2 sin x
= x4 sin x. − 8c1 x2 − 12c2 x3 + 4x3 cos x + 8x2 sin x + 6c1 x2 + 6c2 x3 − 6x2 sin x Hence, y (x) = c1 x2 + c2 x3 − x2 sin x is a solution to the diﬀerential equation for all x ∈ R.
16. We can quickly compute the ﬁrst two derivatives of y (x) = c1 eax + c2 ebx :
y (x) = ac1 eax + bc2 ebx and y (x) = a2 c1 eax + b2 c2 ebx . Substituting these results into the diﬀerential equation yields
y − (a + b)y + aby = a2 c1 eax + b2 c2 ebx − (a + b)(ac1 eax + bc2 ebx ) + ab(c1 eax + c2 ebx )
= (a2 c1 − a2 c1 − abc1 + abc1 )eax + (b2 c2 − abc2 − b2 c2 + abc2 )ebx
= 0. Hence, y (x) = c1 eax + c2 ebx is a solution to the given diﬀerential equation for all x ∈ R.
17. We can quickly compute the ﬁrst two derivatives of y (x) = eax (c1 + c2 x):
y (x) = eax (c2 ) + aeax (c1 + c2 x) = eax (c2 + ac1 + ac2 x)
and
y (x) = eaax (ac2 ) + aeax (c2 + ac1 + ac2 x) = aeax (2c2 + ac1 + ac2 x).
Substituting these into the diﬀerential equation yields
y − 2ay + a2 y = aeax (2c2 + ac1 + ac2 x) − 2aeax (c2 + ac1 + ac2 x) + a2 eax (c1 + c2 x)
= aeax (2c2 + ac1 + ac2 x − 2c2 − 2ac1 − 2ac2 x + ac1 + ac2 x)
= 0.
Thus, y (x) = eax (c1 + c2 x) is a solution to the given diﬀerential equation for all x ∈ R.
18. We can quickly compute the ﬁrst...

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