MATH 23 HOMEWORK #6 PART A SOLUTIONS
Problem 7.1.5. Transform the given initial value problem into an initial value problem for
two first order equations.
1
u00 + u0 + 4u = 2 cos(3t),
4
u(0) = 1,
u0 (0) = −2
Solution. Let x1 = u and x2 = u0 ; then u00 = x02 . In terms of the new variables, we have
1
x02 + x2 + 4x1 = 2 cos(3t)
4
with initial conditions x1 (0) = 1 and x2 (0) = −2. Along with these initial conditions, we
have the system
x01 = x2
1
x02 = − x2 − 4x1 + 2 cos(3t) .
4
Problem 7.1.7. Systems of first order equations can sometimes be transformed into a
single equation of higher order. Consider the system
x01 = −2x1 + x2 ,
x02 = x1 − 2x2 .
(a) Solve the first equation for x2 and substitute into the second equation, thereby
obtaining a second order equation for x1 . Solve this equation for x1 and then determine x2 also.
(b) Find the solution of the given system that also satisfies the initial conditions x1 (0) =
2, x2 (0) = 3.
(c) Sketch the curve, for t ≥ 0, given parametrically by the expressions for x1 and x2
obtained in part (b).
Solution.
(a) Solving the first equation for x2 , we have x2 = x01 + 2x1 . Substituting into
the second, equation, we find
( x01 + 2x1 )0 = x1 − 2( x01 + 2x1 ) ⇐⇒ x001 + 4x01 + 3x1 = 0
which has characteristic equation 0 = r2 + 4r + 3 = (r + 1)(r + 3). Thus the
general solution is
x1 (t) = c1 e−t + c2 e−3t .
Then
x2 = x01 + 2x1 = −c1 e−t − 3c2 e−3t + 2(c1 e−t + c2 e−3t ) = c1 e−t − c2 e−3t .
(b) Imposing the initial conditions, we obtain
c1 − c2 = 3
c1 + c2 = 2,
1
so c1 = 5/2 and c2 = −1/2. Thus
x1 (t) =
5 −t 1 −3t
e − e ,
2
2
x2 (t) =
5 −t 1 −3t
e + e
.
2
2
(c)
Problem 7.1.23. Consider two interconnected tanks similar to those from the previous
problem. Initially, Tank 1 contains 60 gal of water and Q01 oz of salt, and Tank 2 contains
100 gal of water and Q02 oz of salt. Water containing q1 oz/gal of salt flows into Tank 1 at
a rate of 3 gal/min. The mixture in Tank 1 flows out at a rate of 4 gal/min, of which half
flows into Tank 2, while the remainder leaves the system. Water containing q2 oz/gal of
salt also flows into Tank 2 from the outside at a rate of 1 gal/min. The mixture in TAnk
2 leaves it at a rate of 3 gal/min, of which some flows back into Tank 1 at a rate of 1
gal/min, while the rest leaves the system.
(a) Draw a diagram that depicts the flow process described above. Let Q1 (t) and
Q2 (t), respectively, be the amount of salt in each tank at time t. Write down differential equations and initial conditions for Q1 and Q2 that model the flow process.
(b) Find the equilibrium values Q1E and Q2E in terms of the concentrations q1 and q2 .
(c) Is it possible (by adjusting q1 and q2 ) to obtain Q1E = 60 and Q2E = 50 as an
equilibrium state?
(d) Describe which equilibrium states are possible for this system for various values
of q1 and q2 .
Solution.
(a) Let Q1 (t), Q2 (t) be the amount of salt in the respective tanks at time t.
Note that the volume of each tank remains constant. Based on conservation of
mass, the rate of increase of salt, in any given tank, is given by
rate of increase = rate in − rate out .
2
The rate of salt flowing into Tank 1 is
oz
Q2 oz
Q oz
gal
gal
.
rin = q1
3
+
1
= 3q1 + 2
gal
min
100 gal
min
100 min
The rate out is
rout
Q1 oz
=
60 gal
so
gal
Q oz
4
= 1
min
15 min
dQ1
Q
Q
= 3q1 + 2 − 1 .
dt
100
15
Similarly for Tank 2,
Q
3Q2
dQ2
= q2 + 1 −
.
dt
30
100
Thus we have the system
Q2
Q
− 1
100
15
Q
3Q
1
2
Q02 (t) = q2 +
−
30
100
Q01 (t) = 3q1 +
with initial conditions Q1 (0) = Q01 and Q2 (0) = Q02 .
(b) We find the equilibrium values by setting
Q
Q2
− 1
100
15
Q
3Q
1
2
0 = Q02 (t) = q2 +
−
30
100
0 = Q01 (t) = 3q1 +
and solving this system yields Q1E = 54q1 + 6q2 and Q2E = 60q1 + 40q2 .
(c) From the previous part, we see that this amounts to solving
54q1 + 6q2 = 60
60q1 + 40q2 = 50 .
Upon solving this system, we find q1 = 7/6 and q2 = −1/2, but this latter equality
is not physically possible.
(d) Solving the system
Q2E
QE
− 1
100
15
E
Q
3Q2E
0 = q2 + 1 −
30
100
0 = 3q1 +
for q1 and q2 , we find
1 E
1 E
Q1 −
Q
45
300 2
1
3 E
q2 = − Q1E +
Q .
30
100 2
q1 =
3
In order for the result to be physically meaningful, we must have q1 ≥ 0 and
q2 ≥ 0. Then
1 E
1 E
1 E
20 E
1 E
Q1 −
Q2 =⇒
Q2 ≤
Q1 =⇒ Q2E ≤
Q
0 ≤ q1 =
45
300
300
45
3 1
1
3 E
1 E
3 E
10 E
0 ≤ q2 = − Q1E +
Q2 =⇒
Q1 ≤
Q2 =⇒
Q ≤ Q2E
30
100
30
100
9 1
so we must have
20 E
10 E
Q1 ≤ Q2E ≤
Q .
9
3 1
Problem 7.3.20. Find all eigenvalues and eigenvectors of the given matrix.
√ !
3
√1
3 −1
Solution.
√ !
r−
√1 − 3 = r2 − 1 − 3 = r2 − 4 = (r − 2)(r + 2)
det(rI − A) = det
− 3 r+1
=⇒ eigenvalues − 2, 2
For r = −2,




√ !
1
1
1 √
1 √ 
−√3 − 3
−2I − A =
; √
3 ;
3
− 3 −1
0
0
− 3 −1
1
1
=⇒ x1 + √ x2 = 0 =⇒ x1 = − √ x2
3
3


1
1
1
√
x
−
x1
2
=⇒
=
3  = − √ x2 −√3
x2
3
x2
For r = 2,
√ !
√ √
− 3
1 − 3
=⇒ x1 = 3x2
;
2I − A =
0
0
− 3
3
√
√ x1
3x2
3
=⇒
=
= x2
x2
x2
1
1
√
Thus we have eigenvalues −2 and 2 with corresponding eigenvectors
√ 1
3
√
,
.
− 3
1
Problem 7.3.23. Find all eigenvalues and eigenvectors of the given matrix.


3
2
2
 1
4
1 
−2 −4 −1
4
Solution.

r − 3 −2
−2
det(rI − A) = det  −1 r − 4 −1  = r3 − 6r2 + 11r − 6 = (r − 3)(r − 2)(r − 1)
2
4
r+1

= 1, 2, 3. For r = 1,





1
3 1
1
3
1
−2 −2
4 0 
−3 −1 ;  −2 −2 −2  ;  0
0 −2 0
2
4
2
4
2



1 0 1
3 1


0 1 0 
1 0
;
0 0 0
0 0
 

  
−1
− x3
x1





0
x
= x3 0 
=
=⇒ x1 = − x3 , x2 = 0 =⇒
2
1
x3
x3
so we have eigenvalues r

−2

I − A = −1
2

1

0
;
0
For r = 2,





1 2 0
1 2
2
−1 −2 −2
1  ;  0 0 1  =⇒ x1 = −2x2 , x3 = 0
2I − A = −1 −2 −1 ;  0 0
0 0 0
0 0 −1
2
4
3

  
 
x1
−2x2
−2





x2
=⇒ x2 =
= x2 1 
x3
0
0

For r = 3,




0 −2 −2
1
1
1
1
3I − A =  −1 −1 −1  ;  0 −2 −2  ;  0
2
4
4
2
4
4
0
  

x1
0
=⇒ x1 = 0, x2 = − x3 =⇒  x2  = − x3  =
x3
x3




1 1
1 0 0
1 1 ; 0 1 1 
2 2
0 0 0
 
0
x3 −1
1
Thus we have eigenvalues r = 1, 2, 3 with corresponding eigenvectors
     
−1
−2
0
 0  ,  1  , −1 .
1
0
1
Problem 7.3.33. Show that if λ1 and λ2 are eigenvalues of a Hermitian matrix A, and if
λ1 6= λ2 , then the corresponding eigenvectors x(1) and x(2) are orthogonal.
Solution. Since λ1 , λ2 are eigenvalues, then Ax(i) = λi x(i) for i = 1, 2. Since A is Hermitian,
then A∗ = A and its eigenvalues λ1 , λ2 are real. Putting these facts together, then
λ1 h x(1) , x(2) i = hλ1 x(1) , x(2) i = h Ax(1) , x(2) i = h x(1) , A∗ x(2) i = h x(1) , Ax(2) i
= h x(1) , λ2 x(2) i = λ2 h x(1) , x(2) i = λ2 h x(1) , x(2) i .
5
Subtracting, we have
0 = λ1 h x(1) , x(2) i − λ2 h x(1) , x(2) i = (λ1 − λ2 )h x(1) , x(2) i .
Since λ1 6= λ2 , then λ1 − λ2 6= 0. Thus we must have h x(1) , x(2) i = 0, so x(1) and x(2) are
orthogonal.
Problem 7.4.2. In this problem we otline a proof of Theorem 7.4.3 in the case n = 2. Let
x(1) and x(2) be solutions of Eq. (3) for α < t < β, and let W be the Wronskian of x(1) and
x(2) .
(a) Show that
(1)
(1)
(2) (
2
)
dx1
dx1 x
x
1
1 dW
dt + (1)
= dt
(2) .
dx
dx
dt
(2)
x(1)
2 x2 2
2
dt
dt
(b) Using Eq. (3), show that
dW
= ( p11 + p22 )W .
dt
(c) Find W (t) by solving the differential equation obtained in part (b). Use this expression to obtain the conclusion stated in Theorem 7.4.3.
Solution.
(a) This is an immediate consequence of the multilinearity of the determinant, but we provide a computational proof anyway.
Observe that
(
1
)
(2)
dx
(2) dx1 x(1)
1
x
1
1 dt
dt + dx(1) dx(2) 2
2 (1)
(2) x2 dt
x2
dt
(1) 0
(2)
(2) 0
(1)
(1)
(2) 0
(1)
(2)
(2)
(1)
(2)
(1) 0
= x1 (t) x2 (t) − x1 (t) x2 (t) + x1 (t) x2 (t) − x1 (t) x2 (t)
Since
W [ x1 , x2 ](t) = x1 (t) x2 (t) − x1 (t) x2 (t)
then
dW
(1) 0
(2)
(1)
(2) 0
(2) 0
(1)
(2)
(1) 0
= ( x1 (t) x2 (t) + x1 (t) x2 (t)) − ( x1 (t) x2 (t) + x1 (t) x2 (t))
dt
(1) 0
(2)
(2) 0
(1)
(1)
(2) 0
(2)
(1) 0
= x1 (t) x2 (t) − x1 (t) x2 (t) + x1 (t) x2 (t) − x1 (t) x2 (t)
which is the same as (1), establishing the desired equality.
0
(b) By Eq. (3), x(i) = P(t)x(i) for i = 1, 2, so
! !
(1) 0
(1)
p11 p12
x1
x1
=
(1) =
(1) 0
p21 p22
x2
x2
!
! (2) 0
(2)
x1
p11 p12
x1
=
(2) =
(2) 0
p
p
21
22
x2
x2
6
(1)
(1)
!
(2)
p11 x1
(2)
p21 x1
(2)
p12 x2
(2)
p22 x2
!
p11 x1 + p12 x2
(1)
(1)
p21 x1 + p22 x2
+
+
(1)
By part (a) and the multilinearity of the determinant, then
(1)
(1)
(2) (2) dx1
dx1 x
x
1
1 dW
dt + (1)
= dt
(2) dx2 dx2
dt
(2) x(1)
x2 2
dt
dt
(1)
(1)
(2)
(2) (1)
(2)
p
x
+
p
x
p
x
+
p
x
x
x
11 1
12 2 1
1
= 11 1 (1) 12 2
+
(2)
(1)
(1)
(2)
(2) x2
x2
p21 x1 + p22 x2
p21 x1 + p22 x2
0
0
*
*
(1) (2) ( 1 ) (1) (1) (2) (2) (2) x1 x1 x1
x2 x2 x1 x1 x1
= p11 (1) (2) + p12 + p21 + p22 (1) (2) x
x(1) x(2) x(1) x(2) x
x2 x2 2
2
1
1
2
2
(1) (2) x x
= ( p11 + p22 ) 1(1) 1(2) = ( p11 + p22 )W
x
x 2
2
where we are able to cancel the second and third terms because they have repeated
rows.
(c) Using our expression from part (c) and separating variables, then
R
dW
= ( p11 (t) + p22 (t)) dt =⇒ W = ce ( p11 (t)+ p22 (t)) dt .
ln |W | =
W
If c = 0, then W is identically 0 on the interval. Otherwise, if c 6= 0, since the exponential function is always strictly positive, then W never vanishes on the interval.
(Note: There is a part (d) for this problem, but it was not assigned.)
Z
Z
Problem 7.4.6. Consider the vectors
t
(1)
x (t) =
1
and
x
(2)
2
t
(t) =
.
2t
(a) Compute the Wronskian of x(1) and x(2) .
(b) In what intervals are x(1) and x(2) linearly independent?
(c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by x(1) and x(2) ?
(d) Find this system of equations and verify the conclusions of part (c).
Solution.
(a)
W [x
(1)
,x
(2)
t t2
](t) = det
1 2t
= 2t2 − t2 = t2
(b) The Wronskian is nonzero for t 6= 0, so the functions are linearly independent on
(−∞, 0) ∪ (0, ∞).
(c) By Theorem 7.4.3, at least one of the coefficient functions (the entries of P(t)) must
be discontinuous at t = 0.
(d) Letting
2 t
c1
t
c1 t + c2 t2
t t2
x = c1
+ c2
=
=
1
c2
2t
c1 + 2c2 t
1 2t
7
then
0
x =
1 2t
0 2
c1
.
c2
We now try to determine the matrix P(t). Observe that
p11 p12
c1
t t2
Px =
p21 p22
c2
1 2t
Equating x0 and Px, we find
p11 p12
1 2t
t t2
=
.
0 2
p21 p22
1 2t
| {z }
A
We compute
A
−1
=
2/t −1
−1/t2 1/t
and multiplying both sides on the right by A−1 yields
0
1
p11 p12
P=
=
.
p21 p22
−2/t2 2/t
Thus x1 and x2 satisfy the system
0
x =
0
1
2
−2/t 2/t
x
and indeed, we observe that the coefficients of P are discontinuous at t = 0.
Problem 7.4.7. Consider the vectors
x
(1)
2
t
(t) =
2t
x
(2)
t
e
(t) = t ,
e
and answer the same questions as in Problem 6.
Solution.
(a)
h
W x
(1)
,x
(2)
i
t2 et
= det
2t et
= (t2 − 2t)et = t(t − 2)et .
(b) The Wronskian vanishes at t0 = 0 and t0 = 2. Thus the vectors are linearly independent on (−∞, 0) ∪ (0, 2) ∪ (2, ∞).
(c) By Theorem 7.4.3, we must have that one or more of the coefficients of the ODE is
discontinuous at t0 = 0 and t0 = 2; otherwise the Wronskian would not vanish.
(d) Letting
t 2
2 t 2
e
c1
c1 t + c2 et
t e
t
x = c1
+ c2 t =
t =
t
c2
2t
e
2c1 t + c2 e
2t e
then
0
x =
2t et
2 et
8
c1
.
c2
We now try to determine the matrix P(t). Observe that
2 t p11 p12
c1
t e
Px =
t
p21 p22
c2
2t e
Equating x0 and Px, we find
2 t
p11 p12
2t et
t e
.
t = p
p
2 e
2t et
21
22
| {z }
A
We compute

1
1


−

 t(t − 2)
t
(
t
−
2
)
−1
A =


2e−t
te−t 
−
t−2
t−2
−1
and multiplying both sides on the right by A yields


0
1
p11 p12
P=
=  2(1 − t) t2 − 2  .
p21 p22
t(t − 2) t(t − 2)
Thus x1 and x2 satisfy the system


0
1
x0 =  2(1 − t) t2 − 2  x
t(t − 2) t(t − 2)
and indeed, we observe that two of the entries of P are discontinuous at t = 0, 2.

9