MATH 23 HOMEWORK #7 PART A SOLUTIONS

Problem 7.6.3. Consider the system

x =

(2 −51 −2

)x .

(a) Express the general solution of the given system of equations in terms of real-valued functions.

(b) Draw a direction field, sketch a few of the trajectories, and describe the behaviorof the solutions as t→ ∞.

Solution. (a) Letting A be the above matrix, then the characteristic polynomial is

det(A− rI) =∣∣∣∣2− r −5

1 −2− r

∣∣∣∣ = −(2− r)(2 + r) + 5 = r2 − 4 + 5 = r2 + 1

which has roots r = ±i. Then

A− rI =(

2− i −51 −2− i

);(

1 −2− i2− i −5

);(

1 −2− i0 0

)=⇒ ξ1 = (2 + i)ξ2

=⇒(ξ1ξ2

)=

((2 + i)ξ2

ξ2

)= ξ2

(2 + i

1

).

Thus ξ(1) =

(2 + i

1

)is an eigenvector with corresponding solution

x(1) =(

2 + i1

)eit =

[(21

)+ i(

10

)](cos(t) + i sin(t))

=

(2 cos(t)− sin(t)

cos(t)

)+ i(

cos(t) + 2 sin(t)sin(t)

).

Taking real and imaginary parts, we find the general solution

x = c1

(2 cos(t)− sin(t)

cos(t)

)+ c2

(cos(t) + 2 sin(t)

sin(t)

).

(b) The solutions are periodic, hence are remain bounded as t→ ∞.1

Problem 7.6.8. Express the general solution of the given system of equations in terms ofreal-valued functions.

x′ =

−3 0 21 −1 0−2 −1 0

x

Solution. Denoting the above matrix by A, then the characteristic polynomial is

det(A− rI) =

∣∣∣∣∣∣−3− r 0 2

1 −1− r 0−2 −1 −r

∣∣∣∣∣∣ =∣∣∣∣∣∣

1 −1− r 0−2 −1 −r−3− r 0 2

∣∣∣∣∣∣ =∣∣∣∣∣∣1 −r− 1 00 −2 r− 3 −r0 −(r + 3)(r + 1) 2

∣∣∣∣∣∣= −(2r + 3)(2)− r(r + 3)(r + 1) = −r3 − 4r2 − 7r− 6 .

Sincer3 + 4 r2 + 7 r + 6 = (r + 2)(r2 + 2r + 3)

then the eigenvalues are r1 = −2, r2 = −1−√

2i, r3 = −1 +√

2i.

A− 2I =

−1 0 21 1 0−2 −1 2

;

1 1 00 1 20 1 2

;

1 1 00 1 20 0 0

;

1 0 −20 1 20 0 0

=⇒

ξ1ξ2ξ3

=

2ξ3−2ξ3ξ3

= ξ3

2−21

Thus we have the solution

x(1) =

2−21

e−2t .

2

Letting r = −1 +√

2i, then

A− rI =

−3− r 0 21 −1− r 0−2 −1 −r

;

1 −r− 1 0−r− 3 0 2−2 −1 −r

;

1 −r− 1 00 −2r 20 −2r− 3 −r

;

1 −r− 1 0

0 113

r +23

0 −2r− 3 −r

;

1 −r− 1 0

0 113

r +23

0 0 0

;

1 0

13

r− 13

0 113

r +23

0 0 0

Since r = −1 +

√2i, then

r− 1 = −2 +√

2i r + 2 = 1 +√

2i .

=⇒

ξ1ξ2ξ3

=

2−√

2i3

ξ3

−1−√

2i3

ξ3

ξ3

= ξ3

2−√

2i3

−1−√

2i31

.

Taking ξ3 = 3, then we find the eigenvector 2−√

2i−1−

√2i

3

and hence the associated solution

x(2) =

2−√

2i−1−

√2i

3

e(−1+√

2i)t = e−t

2−13

+ i

−√

2−√

20

(cos(√

2t) + i sin(√

2t))

=

2 cos(√

2t) +√

2 sin(√

2t)− cos(

√2t) +

√2 sin(

√2t)

3 cos(√

2t)

e−t + i

−√

2 cos(√

2t) + 2 sin(√

2t)−√

2 cos(√

2t)− sin(√

2t)3 sin(

√2t)

e−t

Taking real and imaginary parts of this second solution yields the general solution

x = c1

2−21

e−2t + c2

2 cos(√

2t) +√

2 sin(√

2t)− cos(

√2t) +

√2 sin(

√2t)

3 cos(√

2t)

e−t

+ c3

−√

2 cos(√

2t) + 2 sin(√

2t)−√

2 cos(√

2t)− sin(√

2t)3 sin(

√2t)

e−t .

3

Problem 7.6.13. Consider the system

x′ =(

α 1−1 α

)x .

(a) Determine the eigenvalues in terms ofα.(b) Find the critical value or values of α where the qualitative nature of the phase

portrait for the system changes.(c) Draw the phase portrait for a value of α slightly below, and for another value

slightly above each critical value.

Solution. (a) Letting A be the above matrix, then the characteristic polynomial is

det(A− rI) =∣∣∣∣α − r 1−1 α − r

∣∣∣∣ = (α − r)2 + 1 = r2 − 2αr +α2 + 1

which has roots

r =2α ±

√4α − 4(α2 + 1)

2=

2α ±√−4

2=

2α ± 2i2

= α ± i .

(b) Sinceα is real, then the eigenvalues are always complex, hence the origin is a spiralpoint. The only critical value is α = 0: when α < 0, the origin is a stable spiralpoint, whenα = 0, the origin is a center, and whenα > 0, the origin is an unstablespiral point.

(c) Forα = −0.3, we compute the following phase portrait

and forα = 0.3, we compute the following phase portait4

Problem 7.6.19. Consider the system

x′ =(

α 10−1 −4

)x .

(a) Determine the eigenvalues in terms ofα.(b) Find the critical value or values of α where the qualitative nature of the phase

portrait for the system changes.(c) Draw the phase portrait for a value of α slightly below, and for another value

slightly above each critical value.

Solution. (a) Letting A be the above matrix, then the characteristic polynomial is

det(A− rI) =∣∣∣∣α − r 10−1 −4− r

∣∣∣∣ = (α − r)(−4− r) + 10 = r2 + (4−α)r− 4α + 10

which has roots

r =α − 4±

√(4−α)2 + 16α − 40

2=

α − 4±√α2 − 8α + 16 + 16α − 40

2

=α − 4±

√α2 + 8α − 242

.

(b) Solvingα2 + 8α − 24 = 0, we find

α =−8±

√64 + 96

2=−8±

√160

2=−8± 4

√10

2= −4± 2

√10 .

Thenα2 + 8α − 24 is> 0 if α < −4− 2

√10

< 0 if − 4− 2√

10 < α < −4 + 2√

10> 0 if − 4 + 2

√10 < α .

Thus for −4− 2√

10 < α < −4 + 2√

10 the matrix has complex eigenvalues, andsince the real part is negative (note that −4 + 2

√10 ≈ 2.32456), this means the

origin is a stable spiral.5

In the other cases, the roots are both real. To determine whether they have thesame sign or opposite signs, we first determine when |α − 4| = ±

√α2 + 8α − 24:

±√α2 + 8α − 24 = |α − 4| =⇒ α2 + 8α − 24 = (α − 4)2 = α2 − 8α + 16

=⇒ 16α = 40 =⇒ α = 5/2 .

Note that 5/2 = 2.5 and −4 + 2√

10 ≈ 2.32456. When −4 + 2√

10 < α < 5/2,then both roots are negative, so the origin is a stable node. When 5/2 < α, thenthe roots are of opposite sign, hence the origin is a saddle point. Finally, whenα < −4 − 2

√10, then both roots are negative, so the origin is a stable node. In

summary,

α < −4− 2√

10 =⇒ origin is a stable node

−4− 2√

10 < α < −4 + 2√

10 =⇒ origin is a stable spiral

−4 + 2√

10 < α < 5/2 =⇒ origin is a stable node

5/2 < α =⇒ origin is a saddle point.

(c) The plots below depict the phase portrait forα = −12,−5, 2.4 and 5.

6

Problem 7.6.28. A mass m on a spring with constant k satisfies the differential equation(see §3.7)

mu′′ + ku = 0 ,

where u(t) is the displacement at time t of the mass from its equilibrium position.

(a) Let x1 = u, x2 = u′, and show that the resulting system is

x′ =(

0 1−k/m 0

)x .

(b) Find the eigenvalues of the matrix for the system in part (a).(c) Sketch several trajectories of the system. Choose one of your trajectories, and

sketch the corresponding graphs of x1 versus t and x2 versus t. Sketch both graphson one set of axes.

(d) What is the relation between the eigenvalues of the coefficient matrix and the nat-ural frequency of the spring-mass system?

Solution. (a) In standard form, the differential equation is

u′′ +km

u = 0 .

Letting x1 = u and x2 = u′, this becomes

x′2 +km

x1 = 0 =⇒ x′2 = − km

x1

so we have the system

x′1 = x2

x′2 = − km

x1

or in matrix form

x′ =(

0 1−k/m 0

)x .

(b) Denoting the matrix above by A, then the characteristic polynomial is

det(A− rI) =∣∣∣∣ −r 1−k/m −r

∣∣∣∣ = r2 + k/m

which has roots r = ±i

√km

.

(c) Since the eigenvalues are pure imaginary, then the origin is a center. For plottingpurposes, we let k = 2 and m = 3.

7

(d) The general solution to the second-order equation is

u(t) = c1 cos

(√km

t

)+ c2 sin

(√km

t

)

which has natural frequency

√km

. We observe that this is exactly the magnitude

of the eigenvalues ±i

√km

, i.e.,

√km

=

∣∣∣∣∣±i

√km

∣∣∣∣∣ .

Problem 7.8.2. Consider the system

x′ =(

4 −28 −4

)x .

(a) Draw a direction field and sketch a few trajectories.(b) Describe how the solutions behave as t→ ∞.(c) Find the general solution of the system of equations.

Solution. (a)8

(b) All points on the line x2 = 2x1 are equilibria. Solutions that pass through points

off this line are all unbounded: solutions(

x1x2

)with trajectories above this line

have x1 → −∞, x2 → −∞ and those below have x1 → ∞, x2 → ∞ as t→ ∞.(c) The characteristic polynomial is

det(A− rI) =∣∣∣∣4− r −2

8 −4− r

∣∣∣∣ = −(4− r)(4 + r) + 16 = r2 − 16 + 16 = r2

which has a double root r = 0. Then

A− 0I = A =

(4 −28 −4

);(

2 −10 0

)=⇒

(ξ1ξ2

)=

(ξ12ξ1

)= ξ1

(12

)so we find one eigenvector with corresponding solution

x(1) =(

12

).

To find a generalized eigenvector, we solve(4 −2 18 −4 2

);(

4 −2 10 0 0

);(

2 −1 1/20 0 0

)=⇒ ξ2 = 2ξ1 − 1/2

=⇒(ξ1ξ2

)=

(ξ1

2ξ1 − 1/2

)= ξ1

(12

)+

(0−1/2

).

We find the generalized eigenvector(

0−1/2

)with corresponding solution

x(2) =(

12

)t +(

0−1/2

).

Thus the general solution is

x = c1x(1) + c2x(2) = c1

(12

)+ c2

[(12

)t +(

0−1/2

)].

9

Problem 7.8.4. Consider the system

x′ =(−3 5/2−5/2 2

)x .

(a) Draw a direction field and sketch a few trajectories.(b) Describe how the solutions behave as t→ ∞.(c) Find the general solution of the system of equations.

Solution. (a)

(b) All trajectories eventually converge to the origin as t→ ∞.(c) The characteristic polynomial is

det(A− rI) =

∣∣∣∣∣∣∣−3− r

52

−52

2− r

∣∣∣∣∣∣∣ = (r + 3)(r− 2) +254

= r2 + r− 6 +254

= r2 + r +14= (r + 1/2)2

which has a double root r = −1/2. We row reduce(−5/2 5/2−5/2 5/2

);(

1 −10 0

)=⇒ ξ1 = ξ2

and find eigenvector(

11

)with corresponding solution

x(1) =(

11

)e−t/2 .

10

We row reduce(−5/2 5/2 1−5/2 5/2 1

);(−5/2 5/2 1

0 0 0

);(

1 −1 −2/50 0 0

)=⇒ ξ1 −ξ2 = −2/5 =⇒ ξ2 = ξ1 + 2/5

=⇒(ξ1ξ2

)=

(ξ1

ξ1 + 2/5

)= ξ1

(11

)+

(0

2/5

)and find generalized eigenvector

(0

2/5

)with corresponding solution

x(2) =(

11

)te−t/2 +

(0

2/5

)e−t/2 .

Thus the general solution is

x = c1x1 + c2x2 = c1

(11

)e−t/2 + c2

[(11

)te−t/2 +

(0

2/5

)e−t/2

].

Problem 7.8.16. Consider again the electric circuit in Problem 26 of §7.6. This circuit isdescribed by the system of differential equations

ddt

(IV

)=

01L

− 1C− 1

RC

( IV

).

(a) Show that the eigenvalues are real and equal if L = 4R2C.(b) Suppose that R = 1 Ω, C = 1 F and L = 4 H. Suppose also that I(0) = 1 A and

V(0) = 2 V. Find I(t) and V(t).

Solution. (a) The characteristic polynomial is

det(A− rI) =

∣∣∣∣∣∣∣∣−r

1L

− 1C− 1

RC− r

∣∣∣∣∣∣∣∣ = r(

r +1

RC

)+

1LC

= r2 +1

RCr +

1LC

which has roots

r =− 1

RC ±√

1R2C2 − 4

LC

2=− 1

RC ±√

L−4R2CLR2C

2.

The roots are equal if and only if the discriminant is 0, which occurs if and only ifL− 4R2C = 0, i.e., L = 4R2C.

(b) With the above values, the matrix A is(

0 1/4−1 −1

). Then A has characteristic

polynomial

r2 + r +14= (r + 1/2)2

11

with double root r = −1/2. We row reduce(1/2 1/4−1 −1/2

);(

1 1/2−1 −1/2

);(

1 1/20 0

)=⇒ ξ1 = −1

2ξ2 =⇒ ξ2 = −2ξ1

=⇒(ξ1ξ2

)=

(ξ1−2ξ1

)= ξ1

(1−2

)and obtain the eigenvector

(1−2

)with corresponding solution

x(1) =(

1−2

)e−t/2 .

We row reduce(1/2 1/4 1−1 −1/2 −2

);(

1 1/2 2−1 −1/2 −2

);(

1 1/2 20 0 0

)=⇒ ξ1 +

12ξ2 = 2 =⇒ ξ2 = 4− 2ξ1

=⇒(ξ1ξ2

)=

(ξ1

4− 2ξ1

)= ξ1

(1−2

)+

(04

)and find the generalized eigenvector

(04

)with corresponding solution

x(2) =(

1−2

)te−t/2 +

(04

)e−t/2 .

Thus the general solution is(IV

)= c1x(1) + c2x(2) = c1

(1−2

)e−t/2 + c2

[(1−2

)te−t/2 +

(04

)e−t/2

].

Imposing the initial conditions, we find(12

)=

(I(0)V(0)

)= c1

(1−2

)+ c2

(04

)=

(c1

−2c1 + 4c2

)hence c1 = 1 and 4c2 = 4 so c2 = 1. Thus the solution of the IVP is(

IV

)=

(1−2

)e−t/2 +

(1−2

)te−t/2 +

(04

)e−t/2

=

(12

)e−t/2 +

(1−2

)te−t/2 .

12