MATH 215/255
Fall 2014
Assignment 2
§1.4, §Exact equations ([Braun’s Section 1.9]), §1.6
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 1.4.8: Solve
x2
1
y 0 + xy = 3 with y(0) = 0.
+1
Answer.
We rewrite the equation as y 0 + x(x2 + 1)y = 3(x2 + 1). We choose the
integrating factor as
R
r(x) = e
x(x2 +1)dx
=e
2
x4
+ x2
4
.
Multiplying the equation by r(x), we obtain
x4
x2
x4
x2
x4
e 4 + 2 · y 0 + e 4 + 2 · (x3 + x)y = e 4 +
x4
x2
d x4 + x2
⇒
[e 4 2 · y(x)] = e 4 + 2 · 3(x2 + 1).
dx
x2
2
· 3(x2 + 1)
Integrating this equation from x0 = 0 to x and using y(0) = 0, we have
Z x 4 2
x4
x2
s
s
+
0
e 4 2 · y(x) − e · y(0) = 3
e 4 + 2 · (s2 + 1) ds
Z x 4 2 0
x2
s
x4
s
⇒ e 4 + 2 · y(x) = 3
e 4 + 2 · (s2 + 1) ds
Z 0x 4 2
4
2
s
s
− x4 − x2
e 4 + 2 · (s2 + 1) ds.
⇒ y(x) = 3e
0
Remark. You need to use definite integral to express this solution, since an initial
condition is imposed.
• 1.4.10: Newton’s law of cooling states that dx
dt = −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k > 0 is a constant. Suppose
that A = A0 cos(ωt) for some constants A0 and ω. That is, the ambient temperature
oscillates (for example night and day temperatures).
a) Find the general solution.
b) In the long term, will the initial conditions make much of a difference? Why or
why not?
Answer.
Write the equation as
dx
+ kx = kA0 cos(ωt).
dt
Choose the integrating factor as
r(t) = e
R
kdt
= ekt .
Multiplying the equation by this r(t), we obtain
ekt x0 (t) + ekt kx(t) = ekt kA0 cos(ωt)
d kt
(e · x(t)) = ekt kA0 cos(ωt).
⇒
dt
We integrate to get
Z
kt
e · x(t) =
kA0 ekt cos(ωt) dt
k
ω
ekt [cos(ωt) + sin(ωt)] + c1 },
2
+ω
k
k
ω
= kA0 2
ekt [cos(ωt) + sin(ωt)] + c
k + ω2
k
= kA0 {
k2
where c = kA0 c1 , by integration by part twice. Thus, the general solution is
x(t) =
ω
k 2 A0
[cos(ωt) + sin(ωt)] + ce−kt .
2
2
k +ω
k
This answers part a).
For part b), note that x(0) =
x(t) =
k 2 A0
k2 +ω 2
+ c. Therefore, we can write
ω
k 2 A0
k
[cos(ωt)
+
sin(ωt)]
+
(x(0)
−
)e−kt .
k2 + ω2
k
k2 + ω2
As limt→∞ x(0)e−kt = 0 for any x(0) ∈ R, the initial condition does not affect the
long term behavior.
• 1.4.102: Solve y 0 + 2 sin(2x)y = 2 sin(2x), for y(π/2) = 3.
Answer.
We choose the integrating factor as
r(x) = e
R
2 sin(2x)dx
= e− cos(2x) .
Multiplying the equation by this r(x), we obtain
e− cos(2x) [y 0 (x) + 2 sin(2x)y(x)] = 2e− cos(2x) sin(2x)
d − cos(2x)
⇒
[e
· y(x)] = 2e− cos(2x) sin(2x).
dx
2
Integrating this equation from x0 = π/2 to x and using y(π/2) = 3, we have
Z x
− cos(2x)
− cos(π)
2e− cos(2s) sin(2s) ds
e
· y(x) − e
· y(π/2) =
π/2
− cos(2x)
⇒ e
· y(x) − 3e
= e− cos(2s) |xπ/2
− cos(2x)
⇒ y(x) = ecos(2x) [3e + e
⇒ y(x) = 2e
cos(2x)+1
− e]
+ 1.
• 1.4.103: Suppose a water tank is being pumped out at 3 L/min. The water tank
starts at 10 L of clean water. Water with toxic substance is flowing into the tank at
2 L/min, with concentration 20 t g/L at time t. When the tank is half empty, how
many grams of toxic substance are in the tank (assuming perfect mixing)?
Answer. Let x(t) be the grams of toxic substance in the tank at time t. Note that
the change of toxic substance during time interval [t, t + 4t] for small 4t is
4x = (the toxic substance flowing in) − (the toxic substance flowing out)
x(t)
= 20t · 2 · 4t − 3 ·
· 4t.
10 − (3 − 2)t
Thus,
4x
x(t)
= 40t − 3 ·
.
4t
10 − t
Letting 4t → 0, we obtain the differential equation:
dx
x
+3
= 40t.
dt
10 − t
Choose the integrating factor as
r(t) = e
R
3
dt
10−t
= (10 − t)−3 .
Multiplying the equation by this r(t), we obtain
(10 − t)−3
⇒
dx
x
+ 3(10 − t)−3
= 40t · (10 − t)−3
dt
10 − t
d
[(10 − t)−3 · x(t)] = 40t · (10 − t)−3 .
dt
We integrate to get
(10 − t)
−3
Z
40t · (10 − t)−3 dt
Z
−1
10
= 40 [
+
]dt,
2
(10 − t)
(10 − t)3
1
= 40[−(10 − t)−1 + 10 · (10 − t)−2 ] + c.
2
· x(t) =
3
Thus,
x(t) = −40[(10 − t)2 − 5(10 − t)] + c(10 − t)3 .
Since x(0) = 0, we have c = 2. It takes 5 minutes for the tank to be half empty.
Therefore,
x(5) = −40[(10 − 5)2 − 5(10 − 5)] + 2(10 − 5)3 = 250.
So the answer is 250 grams.
• Ex 6 (Braun): Find the general solution of the differential equation
y2
dy
− 2yet + (y − et )
= 0.
2
dt
Answer.
Herein, M (t, y) =
y2
2
− 2yet , N (t, y) = (y − et ). Thus,
∂M (t, y)
∂N (t, y)
= y − 2et ,
= −et ,
∂y
∂t
and they are unequal. Let us see if we find an integrating factor µ(t) which is a
function of t. Multiplying µ(t) to the equation, we obtain
µ(t)
y2
dy
− 2µ(t)yet + µ(t)(y − et )
= 0.
2
dt
This equation will be exact if
∂
∂
[µ(t)M (t, y)] = [µ(t)N (t, y)],
∂y
∂t
i.e.,
µ
∂M
dµ
∂N
=N
+µ
.
∂y
dt
∂t
This is only possible when
∂M
∂y
−
∂N
∂t
N
is a function of t alone. Let us check:
∂M
∂y
−
∂N
∂t
N
=
y − 2et − (−et )
= 1.
y − et
Thus, µ(t) satisfies
dµ
=
dt
∂M
∂y
−
N
4
∂N
∂t
µ(t) = µ(t),
(1)
and we take µ(t) = et . That is, if we choose µ(t) = et , then (1) is exact. As
i.e.,
∂φ
y2
y2
= et ( − 2yet ) = et − 2ye2t ,
∂t
2
2
we integrate with respect to t to get
φ(t, y) =
for some function h(y). As
∂φ
∂y
∂φ
∂t
= µM ,
y2 t
e − ye2t + h(y)
2
= µN , we get
∂φ
= et y − e2t + h0 (y) = et (y − et ).
∂y
Thus h0 (y) = 0 and we take h(y) = 0 (if you take h(y) = c, the final answer will be
the same) and
y2
φ(t, y) = et − ye2t = c,
2
is the general solution, i.e., the solution y(t) is the function satisfying
y 2 (t) t
e − y(t)e2t = c.
2
• Ex 8 (Braun): Solve the initial-value problem:
2t cos y + 3t2 y + (t3 − t2 sin y − y)
Answer.
dy
= 0, y(0) = 2.
dt
Herein, M (t, y) = 2t cos y + 3t2 y, N (t, y) = t3 − t2 sin y − y. Thus,
∂M (t, y)
∂N (t, y)
= −2t sin y + 3t2 ,
= 3t2 − 2t sin y,
∂y
∂t
and they are equal. Thus, the equation is exact. As
∂φ
∂t
= M , i.e.,
∂φ
= 2t cos y + 3t2 y,
∂t
we integrate with respect to t to get
φ(t, y) = t2 cos y + t3 y + h(y)
for some function h(y). As N =
∂φ
∂y ,
we get
t3 − t2 sin y − y =
∂φ
= −t2 sin y + t3 + h0 (y).
∂y
5
Thus h0 (y) = −y, and so h(y) = − 12 y 2 . Therefore, the general solution of the differential equation is
1
φ(t, y) = t2 cos y + t3 y − y 2 = c.
2
By using y(0) = 2, φ(0, 2) = −2 = c, thus, c = −2. Therefore, the solution of this
initial-value problem is defined implicitly by the equation
1
t2 cos y + t3 y − y 2 = −2.
2
• Ex 14 (Braun): Determine the constant a so that the following equation is exact,
and then solve the resulting equation:
eat+y + 3t2 y 2 + (2yt3 + eat+y )
Answer.
dy
= 0.
dt
Herein, M (t, y) = eat+y + 3t2 y 2 , N (t, y) = 2yt3 + eat+y . Thus,
∂M (t, y)
∂N (t, y)
= eat+y + 6t2 y,
= 6yt2 + aeat+y .
∂y
∂t
(t,y)
(t,y)
To have ∂M∂y
= ∂N∂t
so that the equation is exact, we need a = 1. So we consider
the exact equation
dy
et+y + 3t2 y 2 + (2yt3 + et+y )
= 0.
dt
As
∂φ
∂t
= M , i.e.,
∂φ
= et+y + 3t2 y 2 ,
∂t
we integrate with respect to t to get
φ(t, y) = et+y + t3 y 2 + h(y)
for some function h(y). As N =
∂φ
∂y ,
2yt3 + et+y =
we get
∂φ
= et+y + 2t3 y + h0 (y).
∂y
Thus h0 (y) = 0, and we take h(y) = 0. So the general solution y(t) is expressed
implicitly by
et+y + t3 y 2 = c.
6
• Ex 18 (Braun): The differential equation f (t)(dy/dt) + t2 + y = 0 is known to have
an integrating factor µ(t) = t. Find all possible functions f (t).
Answer.
Herein, M (t, y) = t2 + y, N (t, y) = f (t). The equation has an integrating
factor µ(t) = t, i.e.,
µ(t)f (t)(dy/dt) + µ(t)(t2 + y) = tf (t)(dy/dt) + t(t2 + y) = 0
[∂µ(t)M (t,y)]
∂y
is an exact equation. Therefore,
=
∂[µ(t)N (t,y)]
.
∂t
Thus,
[∂µ(t)M (t, y)]
d
∂[µ(t)N (t, y)]
= t = (t · f (t)) =
.
∂y
dt
∂t
and
t · f (t) =
Thus, f (t) =
t
2
t2
+ c.
2
+ ct .
• 1.6.4: Take x0 = sin x. a) Draw the phase diagram for −4π ≤ x ≤ 4π. On this
interval mark the critical points stable or unstable. b) Sketch typical solutions of the
equation. c) Find limt→∞ x(t) for the solution with the initial condition x(0) = 1.
Answer.
x
4S
(a)
3S
2S
U
S
S
0
S
: stable
: unstable
2S
3S
4S
(b)
7
dx/dt=sin(x)
5
4
3
2
(i) x(0)=1
(ii) x(0)=−1
(iii) x(0)=4
(iv) x(0)=−4
x(t)
1
0
−1
−2
−3
−4
−5
0
5
10
15
time t
20
25
30
(c) If x(0) = 1, then limt→∞ x(t) = π.
• 1.6.5: Suppose f (x) is positive for 0 < x < 1, it is zero when x = 0 and x = 1, and it
is negative for all other x. a) Draw the phase diagram for x0 = f (x), find the critical
points, and mark them stable or unstable. b) Sketch typical solutions of the equation.
c) Find limt→∞ x(t) for the solution with the initial condition x(0) = 0.5.
Answer.
(a)
x
1
S
0
: stable
: unstable
(b)
8
dx/dt=x*(1−x)
2
0
(i) x(0)=0.5
(ii) x(0)=−0.5
(iii) x(0)=1.2
x(t)
−2
−4
−6
−8
−10
−12
0
5
10
15
time t
20
(c) If x(0) = 0.5, then limt→∞ x(t) = 1.
9
25
30