Mathematics MATB44, Assignment 4, December 2015
Solutions to Selected Problems
7.1 # 22
1. Tank 1 initially contains 30 gal of water and 25 oz of salt, and Tank 2
initially contains 20 gal of water and 15 oz of salt. Water containing 1
oz/gal of salt flows into Tank 1 at a rate of 1.5 gal/min. The mixture
flows from Tank 1 to Tank 2 at a rate of 3 gal/min. Water containing
3 oz/gal of salt also flows into Tank 2 at a rate of 1 gal/min (from the
outside). The mixture drains from Tank 2 at a rate of 4 gal/min, of
which some flows back into Tank 1 at a rate of 1.5 gal/min, while the
rest leaves the system.
Let Q1 (t) be the amount of salt in Tank 1, in ounces (oz).
Let Q2 (t) be the amount of salt in tank 2 at time t, in ounces (oz).
The volume of water in tank 1 is 30 gallons.
The volume of water in tank 2 is 20 gallons.
Water outflow from tank 1:
1.5 + 1.5 gallons/minute flow in
3 gallons/minute flow out
Net outflow = 0.
Water outflow from tank 2:
3 +1 gallons/minute flow in
4 gallons/minute flow out
Net outflow = 0.
So the quantity of water in tank 1 = 30 gallons = constant
Also, the quantity of water in tank 2 = 20 gallons = constant
Amount of salt in tank 1:
Let concentration of salt in tanks 1 and 2 be respectively s1 (t) and s2 (t)
ounces/gallon.
Then
dQ1
= −3 × s1 (t) + 1.5 × s2 (t) + (1oz/gal) × (1.5gal/min)
dt
1
dQ2
= 3s1 (t) − 4s2 (t) + (3oz/gal) × (1gal/min)
dt
Note that Q1 (t) = 30s1 (t), and Q2 (t) = 20s2 (t). So
3
1.5
dQ1
= 1.5 × 1 − Q1 (t) +
× Q2 (t)
dt
30
20
dQ2
3
4
= 3 × 1 + Q1 (t) −
× Q2 (t)
dt
30
20
2. Equilibrium values are achieved when
0 = 1.5 −
dQ1
dt
= 0 and
1 E
3
Q1 + QE
= 1.5
10
40 2
1 E 1 E
Q − Q2
10 1
5
E
E
leading to Q1 = 42 and Q2 = 36.
0=3+
dQ2
dt
= 0 This means
(1)
(2)
E
3. Define x1 (t) = Q1 (t) − QE
1 and x2 (t) = Q2 (t) − Q2 . Then when we
subtract the equation in part (b) from the equation in part (a) we find
1
3
dx1
= − x1 (t) +
× x2 (t)
dt
10
40
dx2
1
1
= + x1 (t) − × x2 (t)
dt
10
5
1. (7.4 # 6)
Consider the vectors
!
x (t) =
t
1
!
x(2) (t) =
t2
2t
(1)
and
(a) Compute the Wronskian of the two vectors. Soln: The Wronskian
of the two solutions is
"
det
t t2
1 2t
2
#
= t2 .
(b) In what intervals are the two solutions linearly independent? Soln:
The Wronskian vanishes at t = 0. Hence the two vectors are linearly
independent on t < 0 and on t > 0.
(c) What conclusion can be drawn a out the coefficients in the system
of homogeneous differential equations satisfied by
x(1)
and
x(2)
?
Soln: At least one of the coefficients is discontinuous at t = 0.
(d) Find this system of equations and verify the conclusions of part (c)
Soln: To find the solution,
"
we look for P =
"
#
p11 p12
p21 p22
t t2
1 2t
which satisfies
#0
"
=
1 2t
0 2
#
"
=P
t t2
1 2t
This is given by
"
P =
1 2t
0 2
"
#
=
1 2t
0 2
"
=
#"
"
−2
t
t t2
1 2t
#−1
2t −t2
−1 t
0
1
−2
−2t
2t−1
#
#
.
2. (7.5 #4) Find the general solution of the equation
d~x
=
dt
"
1 1
4 −2
#
~x
Soln:
Eigenvalues
(1 − r)(−2 − r) − 4 = 0
3
#
r2 + r − 6 = 0
(r + 3)(r − 2) = 0
so r = −3 or r = 2. Eigenvectors r = −3 :
"
1+3 1
4
1
#
~v = 0
If
~v =
v1
v2
then v2 = −4v1 Thus
!
r=2:
So ~x(t) = C1
1
−4
−1 1
4 −4
!
#
where k1 is a constant.
v1
v2
!
=0
!
1
1
so v1 = v2 , ~v = k2
1
−4
= k1
"
!
v1
v2
where k2 is a constant.
!
−3t
e
1
1
+ C2
!
e2t .
The eigenvalues have different signs, so (0, 0) is a saddlepoint.
3. 7.6 #9 Find the solution of the initial value problem
d~x
=
dt
"
1 −5
1 −3
#
~x, ~x(0) =
1
1
!
Soln:
Eigenvalues: (1 − r)(−3 − r) + 5 = 0 r2 + 2r + 2 = 0 r = −1 ± i
Eigenvectors: r = −1 + i :
"
2−i
−5
1
−2 − i
#
~v = 0
(2 − i)v1 = 5v2
~v = k1
4
5
2−i
!
So solution is
"
5
2
~x1 =
!
0
−1
+i
!#
e(−1+i)t
Real solutions: if
~x = (~a + i~b)e(l+im)t
the real solutions are
~u = elt (~a cos mt − ~b sin mt)
~v = elt (~a sin mt + ~b cos mt)
So real solutions are
~a =
5
2
!
0
−1
, ~b =
5
2
!
5
2
!
0
−1
!
−t
~u = e (
−t
~u = e (
!
, l = −1, m = 1
!
cos t −
0
−1
!
sin t +
0
−1
sin t)
cos t)
So the solution is
−t
~x = C1 e (
5
2
!
cos t−
−t
sin t)+C2 e (
Initial condition
~x(0) =
~x(0) = C1
5
2
1
1
5
2
!
sin t+
!
!
+ C2
0
−1
!
So 5C1 = 1 and 2C1 − C2 = 1. So C1 = 1/5, C2 = −3/5 or
−t
~x = e
cos t − 3 sin t
cos t − sin t
So (0, 0) is a spiral point.
a1 , . . . , an , then
5
!
0
−1
!
cos t)
4. 7.8 #1
Find the general solution of the system of equations
d~x
=
dt
"
3 −4
1 −1
#
~x
Soln: Eigenvalues: (3 − r)(−1 − r) + r = 0 r2 − 2r + 1 = 0 (r − 1)2 = 0
so r = 1. Eigenvector:
"
3−1
−4
1
−1 − 1
so
2
1
~v = k
#
~v = 0
!
where k is a constant. So one solution is
~x =
2
1
!
et
Look for a second solution of the form
~x = (~v t + w)e
~ `t
Find
d~x
= (~v e`t + (~v t + w)e
~ `t
dt
= A~x = A(~v t + w)e
~ `t
= (`~v t + Aw)e
~ `t
Hence
(A − `I)w
~ = ~v
"
where A =
#
3 −4
. So
1 −1
"
2 −4
1 −2
6
#
w
~ = ~21
which leads to w1 = 1 + 2w2 if
w1
w2
w
~=
Hence
1
0
w
~=
!
.
!
2
1
+
!
w2
Without loss of generality we may take w2 = 0 so
~x = C1
2
1
!
2
1
t
e + C2
!
t+
1
0
!!
3
2
!
et .
5. 7.8 # 7
Find the solution of the initial value problem
d~x
=
dt
"
1 −4
4 −7
#
~x, ~x(0) =
Soln:
Eigenvalues: (1 − r)(−7 − r) + 16 = 0 (r + 3)2 = 0 so r = −3
Eigenvector:
"
so ~v = k
1
1
4 −4
4 −4
#
~v = 0
!
where k is a constant. One solution is
~x(t) = e−3t~11
A second linearly independent solution is of the form
~x(t) = (~v t + w)e
~ 3t
d~x
= A~x
dt
so
[~v + 3(~v t + w)]e
~ 3t = A(~v t + w)e
~ 3t
7
or
(A − 3)w
~ = ~v
or
"
4 −4
4 −4
#
w
~ = ~11
Hence 4(w1 − w2 ) = 1 or
!
1/4
0
w
~=
so
1
1
~x(t) =
+ w2
!
1/4
0
t+
!
1
1
!!
e−3t
General solution is
~x = C1
1
1
!
e
−3t
3
2
~x(0) =
!
1
1
+ C2
!
= C1
1
1
1/4
0
t+
!
!!
1/4
0
+ C2
e−3t
!
Hence 3 = C1 + C2 /4 and 2 = C1 so C2 = 4. So the solution is
~x = 2
1
1
!
e
−3t
1
1
+4
!
t+
1/4
0
Here (0, 0) is an improper node.
Part B
(1) Eigenvalues r = ±4
Eigenvectors
!
r = +4 : ~v =
3
4
!
r = −4 : ~v =
1
4
This means the solutions are
ξ(t) = C1 e
4t
3
4
!
−4t
+ C2 e
8
1
4
!
!!
e−3t
(b) If C1 = 1, C2 = 0 the solution is the line in the direction
3
4
!
(c) If C1 = 0, C2 = 1 the solution is the line in the direction
1
4
!
(d) If C1 = C2 = 1 the solution is asymptotic to the line
3
4
!
as t → +∞
(e) If C1 = C2 = 1 the solution is asymptotic to the line
1
4
!
as t → −∞
(f) : (0, 0) is a saddle point.
(2) Eigenvalues r = ±2i:
Eigenvectors:
!
~v =
5
−4 + 2i
!
(when the eigenvalue is +2i) and
5
−4 − 2i
(when the eigenvalue is −2i).
(b),(c):
The solution is
~x(t) =
10
−6
!
(cos(2t)) −
0
4
!
sin(2t)
The solution is
C1
5
−4 + 2i
!
e2it + C2
9
4
−4 − 2i
!
e−2it
If C1 and C2 are complex conjugates of each other, we get a real multiple
of
~x(t) =
10
−6
!
(cos(2t)) −
0
4
!
sin(2t)
(d), (e) The trajectories circle around (0, 0) (they are not asymptotic to
any line)
(f) (0, 0) is a center
(3) Eigenvalues r = 6 (repeated root):
Eigenvector
!
1
~
ξ=
2
Solution
~ 6t + C2 e6t (tξ~ + ~η )
C1 ξe
where ~η is a 2-vector which solves the condition
(A − 6I)~η = ξ~
(see text chap. 7.8 Example 2).
(b): line in the direction of ξ~ (through (0, 0))
(c): line parallel to ξ~ containing ~η
NOTE: The class was not taught how to find ~η so they are not responsible
for the material in part B, 3(c)
(d), (e): As t → ±∞ the trajectory is asymptotic to a line in the direction
of ξ
(f) (0, 0) is an improper node.
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