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A First Course in Differential Equations, 3rd ed.
Springer-Verlag, NY (2015)
J. David Logan, University of Nebraska
SOLUTIONS TO ODD-NUMBERED EXERCISES
This supplement contains solutions, partial solutions, or hints to most of theodd-numbered exercises in the text. Many of the plots required in the Exercisesare not displayed, but rather left to the reader. Solutions to the even-numberedproblems are posted on http://www.springer.com/us/book/9783319178516.
Chapter 1 Exercises
� Sec. 1.1.2, page 9
1. (b) x′ = 12
1√6−t2
(−2t) = − tx .
3. Both x = 1/t and x = 1/(t− 2) are solutions.
5. Note x′ = e−x = e− ln(t+C) = 1t+C . This agrees with x′(t) = 1/(t+ C).
7. Substitute x = tm into the differential equation, then cancel the tm termto get 2m = 1, giving m = 1/2.
9. Substitute x = eλt into the differential equation, cancel eλt from each term,and obtain the quadratic function 2λ2 − 5λ− 3 = 0. Thus λ = 3,−1/2.
11. x(t) = (t2 − C)e3t; so x′ = (t2 − C)3e3t + 2t33t = 3x+ 2t33t.
13. Solve 12C0 = C0e
−5730r for t to obtain 0.000121 per year.
15. The decay rate is r = 0.13. Let x0 be the acceptable level of radiation;then x0 = 100x0e
−0.13t. Solving for t gives 35.4 yrs.
17. The population law is p(t)− p0e−µt. We are given p(2) = 90 = 1000e−µ·2.
Solving for µ gives µ = 1.2039 per day.
1
� Sec. 1.1.3, page 15
1. The slope field along both lines x = 0 and x = 4 is zero(horizontal). Notethat x(t) = 0 and x(t) = 4 are obviously solutions.
3. The nullclines for x− = 1− x2 are x = 1 and x = −1. The slope field hasvalue −3 along the curve 1 − x2 = 3, or on x = ±2. The slope field hasvalue +3 nowhere because 1− x2 = +3 has no solutions.
5. 2x2(x− 4√t) = 0 on x = 0 and x = 4
√t. For 0 < x < 4
√t the slope field
is negative; for x < 0 it is negative; and for x > 4√t it is positive.
� Sec. 1.2, page 19
1. x(t) =∫t cos t2 dt = 1
2 sin t2 + C. Then, x(0) = C = 1.
3. Integrate twice to get
x′(t) =
∫−3
√t /dt = −2t3/2 + c1,
x(t) =
∫(−2t3/2 + c1)dt = −4
5t5/2 + c1t+ c2.
The initial conditions give c1 = 4 and c2 = −11/5.
5. By Newton’s second law, mv′ = −F . Integrate and use the initial velocitycondition to get v = −Ft/m+V. Clearly, v = 0 when t = mV/F . Becausev = x′, we have x′ = −Ft/m+ V . Integrate to get
x =F
2mt2 + V t,
where we used x(0) = 0 as the initial condition to evaluate the constantof integration. Next, setting t = mV/F , the stopping time, we get the
distance traveled as X = −mV 2
2F .
7. Integrate the equation twice. The first integration gives
tx′(t) = t+ c1.
So, x′ = 1 + c1t . Integrate again to get
x(t) = t+ c1 ln t+ c2.
Apply the initial conditions to get c2 = 0 and c1 = 1.
9. Use the fact that x′ = 3ye3t + e3ty′ and write the equation in terms of y.
11. Take the derivative using the product rule to obtain
y′ = e−t2et2
− 2te−t2∫ t
0
es2
ds = 1− 2ty.
2
13. Differentiate both sides to get the differential equation
x′(t) = −2e−2t + tx(t).
Also, evaluate at zero to get x(0) = 1.
15. Integrate both sides from 1 to t to get∫ t
1
x′(s)ds =
∫ t
1
5sx2(s)ds+
∫ t
1
ds.
Use the fundamental theorem of calculus for the left side to get
x(t) =
∫ t
1
5sx2(s)ds+ t− 1.
17. By the chain rule,
d
dtI(t, b(t)) =
∂I
∂t(t, b(t))
dt
dt+∂I
∂b(t, b(t))
db
dt
=
∫ b(t)
a
ft(t, s)ds+ f(t, b(t))b′(t).
� Sec. 1.3.1, page 26
1. (a) x(t) = (C + t)2/4.
(b) x(t) = ln(C + 2t)/2.
(c) y(t) = tan(t+ C).
(d) u(t) = 5/2±√
(C − t).
(e) The equation should read x′ = ax+ b. Solution: x(t) = Ceat − b/a.
(f) The implicit solution is 4 ln |Q|+ 12Q
2 = t+ C.
(g) The implicit solution is∫ x
0exp(−r2)dr = t+ C.
(h) y(t) = Ce−rt + a.
3. Apply the initial condition to the general solution given in Problem 1 todetermine C.
5. Separating variables and integrating gives y2 + y = t+ C. Then y(0) = 1forces C = 2. The solution in implicit form is then y2 + y − t − 2 = 0,which is quadratic in y. Solving for y gives
y(t) =1
2
(−1 +
√9 + 4t
).
The plus sign on the radical is taken to satisfy the initial condition. Thesolution exists for t > −9/4.
3
7. x(t) = 1/(1− t2). The interval of existence is (−1, 1).
9. The constants are a = −b = 1/4. The general solution is
x(t) =4Ce4t
1− Ce4t.
11. The general implicit solution, found by separating variables and integrat-ing, is
1
2(ln |x|)2 = 4t+ t2 + C.
The initial condition x(0) = e implies C = 1/2. Thus, (ln |x|)2 = 8t +2t2 + 1. Note x ̸= 0; and the solution is valid as long as 2t2 + 8t+ 1 > 0,or t > −2 +
√7/2.
13. (a) Implicitly differentiating −t3+3x−x3 = C gives −3t2+3x′−3x2x′ = 0.Solving for x′ gives the differential equation. (b) When x = 1, t = 1, weget C = 1. We can plot t vs x and rotate the graph 90 degrees. So,t = (3x− x3 − 1)1/3. (c) Plot the graph.
15. Separate variables and integrate to get
x(t) = 1 + (t2 + C)3.
There is no value of C that gives x(t) = 1 for all t.
17. Separate variables and then use the partial fraction expansion to computethe x integral. Simplify to get
x
K − x= Cert,
and solve for x.
19. Separate variables to get∫dR
R ln(R/K)= −at+ C.
To do the R integral, use the substitution w = ln(R/K), dw = (1/R)dR.Then the general solution can be written
R(t) = KeCe−at
.
21. Let x′ = F (x/t) and let y = x/t, or x = ty. Then x′ = ty′ + y. So thedifferential equation becomes ty′ = F (y)− y, which is separable. We cansolve for y and then use x = ty to obtain the solution. For the example,write the equation as
x′ =4 + 3(x/t)2
2(x/t).
4
Then substitute y = x/t as above and simplify to get
2y
4 + y2dy =
1
tdt.
Integrate and solve for y. Then substitute back to obtain the implicitsolution 4 + (x/t)2 = Ct.
23. Integrate both sides with respect to t to get
xe2t = −e−t + C.
Apply the initial condition x(0) = 3 and solve for x to get x(t) = 4e−2t −e−3t.
25. The governing equation is u′ = −r(t)u. Taking r(t) = at and solving thedifferential equation gives
u(t) = Ce−at2/2.
Notice that the rate is u′ = aCte−at2/2. To find the the time when max-imum rate of conversion occurs, take the derivative of the rate and set itto zero to get tm = 1/
√a.
27. Let h denote the thickness of the ice on the pond. Therefore the differentialequation is h′ = k/h. Solving this equation gives h(t) =
√2kt+ C. Now
h(0) = 0.05, which gives C = 0.0025. In addition, h(4) =√8k + 0.0025 =
0.075. This gives k and therefore the formula for h(t).
� Sec. 1.3.2, page 32
1. The IVP is T ′ = −0.42(T − 350), T (0) = 70. The solution is T (t) =350− 280e−0.42t. Then, T (t) = 200 when t = 1.49 hrs.
3. Let t = 0 occur at 11:30 am. Then T (0) = 94.6 and T (1) = 93.4; also,Te = 70. The solution is
T (t) = 70 + 24.6e−ht.
Then T (1) = 70 + 24.6e−h = 93.4. Solving for h gives h = 0.05. Then
T (t) = 70 + 24.6e−0.05t = 98.6
when t = −3.01. Thus the time of death was about 8:30 am.
� Sec. 1.3.3, page 34
1. The equation C ′ = qV (Ci − C) is separable. The solution to the initial
value problem, with C(0) = C0, is
C(t) = (C0 − Ci)e−qt/V + Ci.
5
3. The volume is variable and given by V (t) = 100− 0.1t The pond emptiesin t = 1000 min. The initial value problem is
[(100− 0.1t)C]′ = 0.5(0.0002)− 0.6C, C(0) = 0.
5. The IVP is 1000C ′ = −2C, C(0) = 5/1000. The solution is C(t) =0.005e−0.002, representing exponential decay.
7. The governing equation is
(V C)′ = qCi − qC − VaC
b+ C.
Constant (equilibrium) solutions occur when (V C)′ = 0 or
qC2 + [aV + bq − qCi]C − qbCi = 0.
This is a quadratic in C. Solve it to obtain the two equilibria and thendetermine the positive root.
9. Note that a′ − b′ = 0 and a′ + c′ = 0. Then a − b = C = a0 − b0, givingb = a−a0+ b0; and similarly, a+ c = a0. Then the first equation becomesa = −ka(a − a0 + b0), which is a logistic type equation. As t → ∞,a→ a0 − b0, b→ 0, and c→ b0.
� Sec. 1.4.1, page 41
1. (a) and (d) are linear.
3. (a) x(t) = t6 + t2
7 + 2942t5 .
(b) x(t) = e−(a+1)eat+tb .
(c) R(t) = 1t ln(1 + t2) + ln 4
t .
(d) N(t) = 92 (e
−t − et) +N0et.
(e) v(t) = [cosx(C − 3) ln(sinx− 1))]/(sinx+ 1).
(f) R(t) = −te−t + (1 + 1/e)t.
5. An integrating factor is e−t2 . The solution is
x(t) =
√π
2et
2
erf(t) + Cet2
.
7. The equation for y is y′ + y = 3t whose solution is y(t) = 3t + Ce−t − 3.Therefore the solution is x(t) = (3/2)t2 − C1e
−t − 3t+ C2.
9. x(t) = Ce−at − ba in both cases.
11. Let x1 and x2 be two solutions; then (x1 + x2)′ + p(t)(x1 + x2) = 2q ̸= q
unless q = 0.
6
13. Let µ(t) = e−∫r(t)dt, which is the integrating factor. Then
N(t) = n0µ−1(t)− hµ−1(t)
∫ t
0
µ(s)ds.
For part (b), r(t) = (1 + sin θ)/5. Thus µ(t) = e−t/5+cos(t)/5.
15. (a) Let y = x2. Then the differential equation becomes y′ = 43ty + 4t
which is linear.
� Sec. 1.4.2, page 46
1. The volume is a function of time, V(t)=60-t. The tank empties when t=60.The governing differential equation is [(60− t)C]′ = 2− 3C, C(0) = 0. Itsimplifies and reduces to a separable equation,
dC
2− 2C=
1
60− tdt.
The solution is C(t) = 1− 13600 (60− t)2.
3. The equation is x′ = kx(N − x) = kNx(1 − x/N). This is a logisticequation whose limiting solution is x(t) → N as t→ ∞.
5. We have M ′ = −aM + I(t), M(0) = m0. This is a linear equation withintegrating factor eat. Solve it to obtain the solution given in the text.
7. Applying Newton’s second law we obtain
v′ +2
t+ 1v = mg, v(0) = 0.
The integrating factor is (t+ 1)2. The solution is
v(t) =mg
3
(t+ 1 +
1
(t+ 1)2
).
9. (c) The differential equation is N ′ = −m(t)N , N(0) = N0. The survivor-ship function is
S(t) =N(t)
N0= e−
∫ t0m(r)dr = e−
∫ t0−((p+1)t/p0t0)dr = e−
p+12p0t0
t2 .
This is a decaying exponential type function. For p = 10 the decay isvery rapid, typical of a fish or insect population. For p = 0 there is slowerdecay, perhaps typical of a human population.
11. The L equation is linear and does not depend on M, so we solve it to get
L(t) = Ce−(µ0+µ)t +λ
µ0 + µ, C = − λ
µ0 + µ
7
Substitute this function into the M equation (linear) to get
M ′ = −δM + µCe−(µ0+µ)t +λµ
µ0 + µ.
Solving gives
M(t) = − µC
µ0 + µ− δe−(µ0+µ)t +
λµ
δ(µ0 + µ).
� Sec. 1.4.3, page 53
1. The governing equation is Q′ + 2Q = 12 with general solution Q(t) =6 + ae−2t. From the initial condition Q(0) = 5 we get a = −1. I(t) =Q′(t) = 2e−2t.
3. From the circuit equation LI ′ + RI + (1/C)Q = 0 we note that Vc =(1/C)Q. So V ′
c = (1/C)Q′ = (1/C)I, and V ′′c = (1/C)I ′. Substituting,
we get LCV ′′c +RCV ′
c + Vc = 0..
7. The equation is LQ′′ + (1/C)Q = 0. Substitute Q = A cosωt into theequation to get, after cancelation of the cosine terms, −Lω2 + 1/C = 0,giving the frequency ω =
√1/LC. The amplitude A cancels out; it could
be determined by initial conditions.
9. (a) By separating variables, x(t) = C exp(∫p(t)dt).
(b) With p(t) = −2t, the previous solution gives x(t) = Ce−t2 . The initialcondition x(1) = 2 gives C = 2e.
(c) The problem on the interval [0, 1] is x′ = −2x, x(0) = 5 with solutionx1(t) = 5e−2t. The problem on the interval [1, 2] is x′ = −x2, with generalsolution x2(t) = 1/(t + C). The constant C is determined by matchingthe solution x1(t) and x2(t) at t = 1. Thus, 5e−2 = 1/(1 + C), givingC = (e2 − 5)/5.
� Sec. 1.5.1, page 63
1. The model equation is
N ′ = 0.2N(1−N/40)− 1.5.
The right side of the equation is a quadratic and has two positive rootsN = 10, 30 (the equilibria). It plots as a concave down parabola, so thesmaller equilibrium is unstable and the larger one is stable. If N(0) < 10the population becomes extinct; if N(0) > 10 the population tends toN = 30, the most likely scenario.
3. (a) The equilibria are x = 0, 1. Note fx(0) = 2 > 0, so x = 0 is unstable;and, fx(1) = −(1− e−2 < 0, so x = 1 is stable.
8
(c) The right side is zero when R2 − 3R+1 = 0, or R = 32 ±
√52 . So there
are two positive equilibria. The smaller one is unstable and the larger oneis stable. (Either make a plot of R′ vs. R or test the analytic condition.)
(d) Plot 1/z + a2 (hyperbola) and ln z on the same z′ vs z axes. There isa single intersection at the equilibrium. It is stable (subtract the curvesgraphically.)
5. The equilibria are x = −6, x = 6, x = a > 6. The phase line: x′ > 0 forx < −6, x′ < 0 for −6 < x < 6, and x′ > 0 for 6 < x < a, x′ > 0 for x > a.Thus, the three equilibria are stable, unstable, and unstable, respectively.If x(0) = 8, then x(t) moves to the right toward the equilibrium a, andperhaps further if a small perturbation is present at a.
7. Setting x′ = x(1 − x) − h = 0 gives a quadratic for x with roots x =12 ± 1
2
√1− 4h. For h¿1/4 there are no equilibrium. When h < 1/4 there
are two positive equilibria; the small one is unstable and the large one isstable. The population goes to extinction when x(0) < 1
2 − 12
√1− 4h.
9. (a) The predation rate P (N) is a Type 3 Holling response (see a genericplot on p. 278, Figure 5.14).
(b) Note that N = bn and dN/dt = adn/dτ . Substitution into the differ-ential equation gives, after simplification,
dn
dτ= n
[ρ(1− n
s
)− n
n2 + 1
]= nF (n),
where ρ ≡ rb/a and s ≡ K/b. Generally this equation can have four roots(equilibria) and F (n) can have up to three real roots. To determine theroots of F (n) geometrically, plot both terms, ρ
(1− n
s
)and n
n2+1 sepa-rately, and determine any intersection points. The first is a straight linejoining (0, ρ) to (s, 0), and it depends on the two parameters ρ and s. Thelatter is a fixed curve. See Figure 1. As the parameters s and ρ change,the straight lines moves and the equilibria change with accompanying bi-furcations. The figure shows a method to determine the stability of thevarious equilibria.
11. Substituting S=N-I into the I equation gives, upon simplification,
I ′ = (aN − r)(
(1− I
(aN − r)/a
),
which has the form of the logistic equation. Therefore, the limiting valueof the number infected is (aN − r)/a provided we assume that aN > r.
13. The model is
P ′ = α
(k
P−mP
),
9
n
n�
fixed curve
line 1
line 2
line 3
n
stable
s
r
Figure 1: The fixed curve n2
n2+1 and three lines ρ(1− n
s
)for three different sets
of parameters s and ρ. Intersections are shown. Line 1 has 1 intersection, Line2 has 2, and Line 3 has 3 (the third, not shown) will occur for a large value ofs. Drawing the phase line shows the stability for each equilibrium point; e.g.,the figure shows that the equilibrium originating from Line 1 is stable.
where k and m are constants of proportionality. It is reasonable that thedemand should decrease with price, and supply should increase with price.The equilibrium is Pe =
√k/m and it is stable.
� Sec. 1.5.2, page 70
1. (b) The equilibria are along the straight lines x = 1 and x = h in thehx plane. To check stability note that fx(x, h) = 2x − h − 1. Note thatfx(1, h) = 1 − h < 0 for h > 1 (stable) and fx(1, h) = 1 − h > 0 forh < 1 (unstable). Similarly, fx(h, h) = h − 1 < 0 for h < 1 (stable) andfx(1, h) = h− 1 > 0 for h > 1 (unstable).
3. The equilibria are y = 0 and y = ±√λ− 1. In the λy plane these plot as
a horizontal line (y axis) and a parabola through (1, 0) that opens to theright. Note that fy(y, λ) = λ − 1 − 3y2. Hence, fy(0, λ) = λ − 1 < 0 forλ < 1. So the λ axis is stable to the left of the bifurcation point (1, 0), andunstable to the right. Both branches of the parabola are stable becausefy(±
√λ− 1, λ) = −2(λ− 1) < 0.
5. Note that h, r, and P are nonnegative. By factoring out P we see thatequilibria are P = 0 and P = K
r (r−h). The latter equilibrium exists onlywhen r > h. The bifurcation diagram in the hP plane consists of a stablestraight line of equilibria from (0,K) to (r, 0) adjoined to an unstablesegment of equilibria (h, 0), h > r, along the h axis. So, as h increases
10
from a small value, the equilibrium population decreases and disappearsat h = r, becoming extinct. To check stability of the first branch, notethat
f(K(r − h)/r, h) = −r + h < 0, h < r.
9. Not that y′ = 0 when b = e−y2
. On a set of yb coordinates this is a positivebell-shaped curve. We can obtain a plot on a by coordinate system byrotating the curve through a 45 degree angle, giving a bell-shaped curvein the b > 0plane. To check stability of the two branches, note that
fy(y, b) = 2ye−y2
< 0 for y < 0.
So the lower branch of the bell-shaped curve is stable. By a similar cal-culation, the upper branch is unstable.
11. In the resistor the rate of energy change of the total energy is equal tothe rate of ohmic (joule) heating R(T )I2 minus the rate heat is lost tothe environment, given by −h(T − Te), which is Newton’s law of cooling.The total heat energy in the resistor is given by its mass m (kg) times itsspecific heat c (joules/kg deg) times its temperature T (deg). Thus, thetemperature dynamics are described by
d
dt[mcT ] = R(T )I2 − h(T − Te) =
V 2
R(T )− h(T − Te).
(The reader should check the units of each term to see that they agree.)To check equilibria in the case R(T ) is positive and increasing, note thatthere is always a value of T for which
V 2
R(T )= h(T − Te).
This is because the left side is a positive decreasing function that ap-proaches zero as T → ∞ and the right side is a positively-sloped straightline that goes to infinity.
� Sec. 1.5.3, page 76
1. The functions f(t, x) = (t2 + 1)x − t and fx(t, x) = t2 + 1 are bothcontinuous in the entire tx plane, and thus the IVP has a unique solutionfor any initial condition.
3. The function f(t, x) = ln(t2 + x2) is continuous everywhere except theorigin (0, 0). The same is true for fx(t, x) = 2x/(t2 + x2). Therefore theIVP has a unique solution for any initial condition x(t0) = x0 provided(t0, x0) ̸= (0, 0).
5. (a) The function f(t, x) = (2 + t2)/x(3 − x2) as well as fx(t, x) are con-tinuous everywhere except along the lines x = 0 and x = ±
√3. Thus, the
IVP has a unique solution for any initial point not on those lines.
11
(b) The function f(t, x) = (t−x)/(2t+5x) as well as fx(t, x) are continuouseverywhere except along the line x = −2t/5. Thus, the IVP has a uniquesolution for any initial point not on that line.
7. Note that the function f(x, t) =√x is continuous at (0, 0) and thus it has
a solution with initial condition x(0) = 0. However, fx(t, x) = 1/2√x is
not continuous at (0, 0) and the existence-uniqueness theorem hypothesesdo not hold. So there may be many solutions.
12
Chapter 2 Exercises
� Sec. 2.1, page 84
1. Balance gravity and spring force to get mg = kL, where L is the elonga-tion. Then k = mg/L = 0.3 · 9.8/0.05 = 58.8 (N/m) .
3. In the y system, my′′ = −ky+mg. But y = x+∆L. Then y′′ = x′′ so thatmx′′ = −k(x+∆L)+mg = −kx−k∆L+mg = −kx. (The last two termscancel.) When damping is present, the y equation is my′′ = −kyγy′+mg.Proceed as above to show mx′′ = −kx− γx′ +mg.
� Sec. 2.2.2, page 90
1. (a) x(t) = c1e2t + c2te
2t.
(b) x(t) = c1e2t + c2.
(c) x(t) = c1e−t + c2te
−t.
(d) x(t) = c1e−t + c2e
−3t.
3. Using 1(d), x(0) = c1 + c2 = −1, and x′(0) = −c1 − 3c2 = 2. Add thesetwo equations to get c1 = c2 = −1/2.
5. a(c1x1+ c2x2)′′+ b(c1x1+ c2x2)
′+ c(c1x1+ c2x2) = c1(ax′′1 + bx
′1+ cx1)+
c2(ax′′2 + bx′2 + cx2) = 0 + 0 = 0.
7. From the formula for x(t), we have x(0) = 1 and x′(0) = −3 + 2 = −1.
� Sec. 2.2.3, page 94
1. (a) x(t) = e−t/2(c1 cos(√15t/2) + c2 sin(
√15t/2)).
(b) x(t) = e2t(c1 cos(√2t) + c2 sin(
√2t)).
(c) x(t) = c1 cos 3t+ c2 sin 3t.
(d) x(t) = c1e−√12t + c2e
√12t.
(e) x(t) = e−3t/4(c1 cos(√15t/4) + c2 sin(
√15t/4)).
(f) x(t) = c1e−t/3 + c2e
−4t/3.
3. From the solution, x(0) = 3 and x′(0) = 4. The characteristic polynomialhas roots ±4i, and therefore must be λ2+16 = 0. The differential equationis x′′ + 16x = 0.
5. From the form of the solution the eigenvalues are −2± 4i. So the charac-teristic polynomial must be (λ−(−2+4i))(λ−(−2−4i)) = 0. To multiplyout, it is helpful to see that this is of the form (λ− a)(λ− a) = 0, or
λ2 − (a+ a)λ+ aa = 0,
where a = −2 + 4i and a = −2− 4i. Clearly, a+ a = −4 and aa = 20. Sothe characteristic polynomial is λ2+4λ+20 = 0. The differential equationis x′′ + 4x′ + 20x = 0.
13
7. The eigenvalues must be complex and given by −1/2± 3i. From Exercise5, the characteristic polynomial is λ2 + λ+ 1
4 +9 = 0, and the differentialequation is x′′ + x′ + (37/4)x = 0.
� Sec. 2.2.4, page 99
1. The general solution is
x(t) = e−t/2(c1 cos(√3t/2) + c2 sin(
√3t/2)).
Initial conditions give c1 = 1 and c2 =√3. Therefore A = 2 and ϕ =
arctan(√3) = π/3. Thus the phase amplitude form is
x(t) = 2e−t/2 cos(√3 t/2− π/3).
3. The eigenvalues are
λ = −1
2(−1±
√1− 4k.
The eigenvalues can never be purely imaginary, so the system will neveroscillate. If k > 1/4 the eigenvalues are complex and the system willoscillate and decay. If k ≤ 1/4 the roots are real and the system willdecay.
5. The eigenvalues are
λ =1
2[−3δ ±
√9δ2 − 4κ].
The system is critically damped when 9δ2 − 4κ = 0, or κ = 9δ2/4. Thisplots as a parabola in the δκ plane.
� Sec. 2.3.1, page 110
1. (a) xp(t) = At3+Bt2+Ct+D. (b) xp(t) = A cos t+B sin t. (c) xp(t) = A.(d) xp(t) = e3t(Bt2 +Ct+D). (e) xp(t) = A cos 7t+B sin 7t. (f) xp(t) =e2t(A cos t+B cos t) + (Ct2 +Dt+ E).
3. The eigenvalues are
λ =b
2± 1
2
√b2 − 4.
For −2 < b < 1 we get complex roots, and for b ≤ −2 we get real roots .For the complex case, the homogeneous solution is
xh(t) = eb/2[(c1 cos
√4− b2 t+ c2 sin
√4− b2 t
]. (−2 < b < 1)
The particular solution has the form xp(t) = A cos t+B sin t.
In the real case, the homogeneous solution has the form
xh(t) = c1e(b/2+
√b2−4/2)t + c2e
(b/2−√b2−4/2)t. (b < −2)
14
The same form of the particular solution holds in this case.
For b = −2 there are two real equal roots, b = −1,−1. The homogeneoussolution has the form
xh(t) = e−t(c1 + c2t).
The particular solution remains the same.
5. The homogeneous solution is xh(t) = c1 + c2e2t. Because 1 is a basic
solution, we take the particular solution to be xp(t) = At (not just A).Substituting this into the differential equation gives A = −2. So thegeneral solution is
x(t) = c1 + c2e2t − 2t.
Now apply the initial conditions to determine c1 and c2.
7. Use the double angle formula to write the right side of the differentialequation as sin2 t = 1
2 (1 − cos 2t). Then the particular solution will havethe form Ip = A+B cos 2t+ C sin 2t.
9. The model is 5x′′+2x = 10. The general solution is x(t) = c1 cos(√2/5t)+
c2 sin(√
2/5t) + 5. Apply the initial conditions x(0) = 15, x′(0) = 4 todetermine c1 and c2.
11. The differential equation is 2Q′′+16Q′+50Q = 110. A particular solutionis Qp(t) = 2.2. The homogeneous solution is Qh(t) = e−4t(c1 cos 3t +c2 sin 3t). The initial conditions are Q(0) = 5, Q′(0) = 0. The generalsolution is
Q(t) = e−4t(c1 cos 3t+ c2 sin 3t) + 2.2.
The transient part of the solution is the oscillating, decaying homogeneoussolution; it decays away and leaves the transient, constant solutionQp(t) =2.2 We can find the constants by applying the initial conditions; we getc1 = 2.8 and c2 = 3.73.
� Sec. 2.3.2, page 114
3. I(t) = c1 cos 4t+ c2 sin 4t+At cos 4t+Bt sin 4t.
5. The equation is is 4x′′ + kx = 412 sin 5t. Pure resonance occurs when thenatural frequency
√k/4 equals the forcing frequency, 5. Thus k = 100.
7. The general solution has the form
x(t) = c1 cosωt+ c2 sinωt+cosβt
ω2 − β2.
Applying the initial conditions gives c2 = 0 and c1 = −1/(ω2−β2). Thus,
x(t) =cosβt
ω2 − β2(cosβt− cosωt) .
A plot of the solution is shown in Figure 2.9.
15
� Sec. 2.4.1, page 120
1. (a) x(t) =√t(c1 cos
(√32 ln t
)+ c2 sin
(√32 ln t
)).
(b) x(t) = c1t√5 + c2t
−√5.
(c) x(t) = c1t−1+
√2 + c2t
−1−√2.
(d) x(t) = c11t + c2
1t2 .
(e) x(t) = t4(c1 + c2 ln t).
(g) x(t) = 2 ln t.
3. (a) Consider the Cauchy-Euler equation at2x′′ + btx′ + cx = 0, wherex = x(t). Let τ = ln t or t = eτ . Letting X(τ) = x(eτ ), the chain rulegives
x′ =dx
dt=dX
dτ
dτ
dt=
1
t
dX
dτ
and
x′′ =d
dt
dx
dt=
d
dt
(1
t
dX
dτ
)=
1
t2d2X
dτ2− 1
t2dX
dτ.
Substituting these expressions into the Cauchy-Euler differential equationgives, after simplification,
ad2X
dτ2+ (b− a)
dX
dτ+ cX = 0,
which is a constant coefficient equation for X(τ).
(b) Take t2x′′ + x = 0 from Problem 1a. The solution is given in Exercise1a, above. In terms of X(τ), the equivalent constant coefficient differentialequation becomes
d2X
dτ2− dX
dτ+X = 0.
The characteristic polynomial has eigenvalues λ = 12 ± i
√32 . Therefore the
solution is
X(τ) = eτ/2(c1 cos
√3
2τ + c2 sin
√3
2τ).
Making the substitution τ = ln t gives the same solution as in Exercise 1a.
16
� Sec. 2.4.2, page 124
1. (a) A fundamental solution set for the homogeneous equation is {cos t, sin t}.The Wronskian is W (t) = 1. By the variation of parameters formula
xp(t) = − cos t
∫sin t tan t
1dt+ sin t
∫cos t tan t
1dt
= − cos t
∫sin2 t
cos tdt+ sin t
∫sin tdt
= − cos t
∫1− cos2 t
cos tdt− sin t cos t
= − cos t
∫(sec t− cos t)dt− sin t cos t
= − cos t(ln | sec t+ tan t|+ sin t− sin t cos t
= −cost ln | sec t+ tan t| − 2 sin t cos t.
(c) The basic solutions to the homogeneous equation are et and e−t, andthe Wronskian is W (t) = −2. The variation of parameters formula is,after some simplification,
xp(t) =1
2
(etEi(−t)− e−tEi(t)
),
where Ei(t) =∫∞t
e−r
r dr is a special function called the exponential inte-gral.
(d) xp(t) =t3
4 .
5. Note the typo: the corrected equation is Lϕ ≡ ϕ′′+pϕ+qϕ = 0. The initialconditions are ϕ(0) = 0, ϕ′(0) = 1. We verify the particular solution
xp(t) =
∫ t
a
ϕ(t− s)f(s)ds
to Lx = f(t). By Leibniz rule
x′p(t) =
∫ t
a
ϕt(t− s)f(s)ds+ ϕ(0)f(t) =
∫ t
a
ϕt(t− s)f(s)ds,
and
x′′p(t) =
∫ t
a
ϕtt(t− s)f(s)ds+ ϕt(0)f(t) =
∫ t
a
ϕtt(t− s)f(s)ds+ f(t).
Adding the last three equations gives x′′p(t) + px′p(t) + qxp(t) = f(t).
17
7. The basic solutions to the homogeneous equation are et and e−t, and theWronskian is W (t) = −2. The variation of parameters formula is, aftersome simplification,
xp(t) =et
2
∫1
1 + etdt− e−t
2
∫e2t
1 + etdt
=et
2
(t− ln(1 + et)
)− e−t
2
(et + ln(et + 1)
).
� Sec. 2.4.3, page 125
3. (a) Substitute x = tβ into the differential equation to obtain, after cance-lation,
β(β − 1)t−2 − β(β − 1) + 2(1− β) = 0.
The first term cannot be balanced, so either β = 0 or β = 1. The onlychoice that makes the other terms vanish is β = 1. Thus x(t) = t is asolution.
(b) To find another solution set x(t) = tv(t) and substitute into the dif-ferential equation. It simplifies to (t − t3)v′′ + 2(1 − 2t2)v′ = 0. Lettingw = v′, we get the first-order separable equation
dw
w= 2
2t2 − 1
t− t3dt.
Proceed with a partial fraction expansion of the right side and integrateto solve for w. Then v(t) =
∫w(t)dt.
5. Let x(t)=tv(t) and substitute into the differential equation to obtain, aftersimplification, t3v′′ + (2t2 − t − 2)v′ = 0. With w=v’ we get a separableequation
dw
w=
(−2
t+
1
t2+
2
t3
)dt.
Integrating and taking the exponential of both sides gives
w(t) =1
t2e−(1/t+1/t2).
Then
x(t) = t
∫1
t2e−(1/t+1/t2)dt.
7. Note the typo. Let x(t) = v(t)y(t), where y is a solution. Substituting intothe differential equation gives (after using the fact that y′′+py′+qy = 0),yv′′ + 2y′v′ + pyv′ = 0. Letting v′ = z, we get
z′ +
(2y′
y+ p
)z = 0.
18
This is a first-order separable
dz
z= −
(2y′
y+ p
)dt.
Integrate both sides to get
ln z = −2 ln y −∫p(t)dt.
Exponentiating,
z(t) =e−
∫p(t)dt
y2.
Therefore,
x(t) = Cy(t)
∫e−
∫p(t)dt
y(t)2dt.
9. (a) Let x(t) = exp∫y(t)dt. Then
x′(t) = exp
∫y(t)dty(t), x′′(t) = exp
∫y(t)dty′(t) + exp
∫y(t)dty2(t).
Substituting into the differential equation x′′ + px′ + qx = 0 gives y′ +y2 + py + q = 0 .
(b) Let y = x′
x . Then
y′′ =x′′ − (x′)2
x2.
Substituting into the Riccati equation gives x′′ + px′ + qx = 0.
(c) Use part (b) to write the Riccati equation as the linear equation
x′′ − 3
tx′ = 0.
Letting z = x′ gives z′ − (3/t)z = 0. This equation is separable andeasily gives z = c1t
3. Thus x(t) =∫z(t)dt = Ct4. Finally, we have y(t) =
x′(t)/x(t) = 4/t.
(d) Let y′+ay = −by2+c and make the substitution y = x′/bx. Substitut-ing into the differential equation and simplifying leads to x′′+ax′−bcx = 0,a second-order linear equation. It is solved by standard methods and theny is obtained by the substitution.
(e) The chemical reactor equation can be written as in part (d) as
C ′ +q
VC = −kC2 +
q
VCi.
Make the substitution x = C ′/kC to transform the equation into a linearequation
x′′ +q
Vx′ − kCiq
Vx = 0.
.
19
� Sec. 2.5, page 130
1. (a) The characteristic equation is λ3 + λ = 0, giving λ = 0,±i. So thegeneral solution is x(t) = c1 + c2 cos t+ c3 sin t.
(b) The characteristic equation is λ4 + λ = λ(λ+ 1)(λ2 − λ+ 10). So thegeneral solution is
x(t) = c1 + c2e−t + et/2(c3 cos(
√3t/2) + c4 sin(
√3t/2)).
A particular solution is xp(t) = t.
(c) x(t) = c1 + c2t+ c3 cos t+ c4 sin t.
(d) The characteristic equation is λ3 − λ− 8 = 0. To find the roots use acalculator or MATLAB. You get 2.1663, −1.0832±1.5874i. So the generalsolution is
x(t) = c1e2.1663t + e−1.0832t(c2 cos 1.5874t+ c3 sin 1.5874t).
(f) The characteristic equation is λ3 − 8 = 0, or (λ− 2)(λ2 + 2λ+ 4) = 0.Therefore λ = 2, −1± i
√3. So,
x(t) = c1e2t + e−t(c2 cos
√3t+ c3 sin
√3t).
3. The eigenvalues must be 0, 0, 5± 2i. So the characteristic equation is
λ2(λ− (5 + 2i))(λ− (5− 2i)) = λ2(λ2 − 10λ+ 29 = 0).
Therefore the differential equation is x′′′′ − 10x′′′ + 29x′′ = 0.
� Sec. 2.6, page 135
1. The temperature in the bar is u(x) = 30 − 12x. So u(12) = 24. The heat
is leaving at the right end at the rate −Ku′(40) = −K2 [energy/area ×
time].
3. The equation is −(K(x)u′)′ = 0. Integrating once gives K(x)u′ = C1, oru′ = C1/K(x). Integrating again gives
u(x) = C1
∫ x
0
1
K(s)ds+ C2.
When u(0) = u(L) = 0, we get u(x) ≡ 0. When K(x) = (x + 1), thesolution is u(x) = C1 ln(x + 1) + C2. If the left end is held at u(0) = 0and the right end is insulated, or K(L)u′(L) = 0, the temperature in thebar is u(x) = 0. (There are no sources.)
5. Integrating gives uu′ = C or 12 (u
2)′ = C. Integrating again then givesu2 = c1x + c2, or u(x) =
√c1x+ c2. Applying the boundary conditions
leads to u(x) =√4πx.
20
7. It is straightforward to check that there are no nonpositive eigenvalues.Therefore, take λ = p2. Solving u′′ + p2u = 0 gives u(x) = A cos px +B sin px. Next, u′(0) = 0 implies B = 0, so u(x) = A cos px. Then,u(1) + u′(1) = A cos p− pA sin p = 0, giving
tan p =1
p.
This determines the eigenvalues λn = p2n, n = 1, 2, 3, . . .. We can find thenumerical values of the roots pn of the equation using a nonlinear equationsolver. Or, we can find them graphically by plotting tan p and 1/p on thesame set of coordinates.
9. Note the typo. The Dupuit–Forschheimer equation should be
K
2
(h2)′′
= −q.
This equation can instantly be integrated twice to get
h(x)
√− q
Kx2 + c1x+ c2.
Applying the two boundary conditions h(0) = h0 and h(L) = h1 givesc2 = h20 and c1 = qK
L + 1L
(h21 − h20
).
21
Chapter 3 Exercises
� Sec. 3.1, page 144
1. X(s) =∫ 2
1e−stdt = − 1
se−st|21 = −1
se−2s + 1
se−s.
3. Check your answers in the Table.
5. (a) It follow by definition that
F (s− a) =
∫ ∞
0
f(t)e−(s−a)tdt =
∫ ∞
0
f(t)eate−stdt.
(b) Make the substitution r = t− a, dr = dt to get∫ ∞
0
H(t− a)f(t− a)e−stdt =
∫ ∞
a
f(t− a)e−stdt
=
∫ ∞
0
f(r)e−s(r+a)dr = e−as
∫ ∞
0
f(r)e−srdr.
7. By Exercise 6,
L[t2H(t− 1)] = e−sL[(t+1)2] = e−sL[t2 +2t+1] = e−s
(2
s3+
2
s2+
1
s
).
9. (a) 7e−2t. (b) 3− 2√6sin
√6t. (c) 2
t e5t. (d) 7H(t− 4).
13. Make the substitution r = r/a, dt = dr/a.
15. Taking the transform and using linearity,
F (s) =∞∑0
(−1)n1
se−ns =
1
s
∞∑0
(−e−s
)n=
1
s
1
1 + e−s.
17. (a) Let u = tn, dv = e−tdt. Then du = ntn−1, v = − − e−t. Usingintegration by parts,
Γ(n+ 1) =
∫ ∞
0
tne−tdt = −tne−t|∞0 + n
∫ ∞
0
tn−1e−tdt = nΓ(n).
(b) Make the substitution t = r2, dr = 2rdr. Then
Γ(1/2) =
∫ ∞
0
e−tt−1/2dt =
∫ ∞
0
e−r2r−12rdr = 2
∫ ∞
0
e−r2dr =√π.
(c) Make the substitution r = st, dr = sdt. Then∫ ∞
0
tae−stdt =
∫ ∞
0
(rs
)ae−r 1
sdr =
Γ(a+ 1)
sa + 1.
22
� Sec. 3.2, page 156
1. We get a = −1, b = −1 and c = 1. In general,
L−1
(a
s+
b
s2+
c
s− 1
)= a+ bt+ cet.
3. (a) (8/9)e−8t + et/9.
(b) [−2 cos(3t)− 5 sin(3t))/3]e−t.
(c) t3e5t/3.
(d) 12H(t− 1)e2(t−1) − 1
2 .
(e) 7 cos 2t+ 12 sin 2t.
(f) 32
√2/7
√7/2
s2+√
7/22 .
(g) 48!e
3tt8.
(h) e3t(2 cos 2t+ (15/2) sin 2t).
(i) This problem is misstated.
(j) H(t− 1)e1/2−t/2[cos(√3(t− 1))/2)− 1
3
√3 sin(
√3(t− 1))/2)].
(k) t− (t− 2)H(t− 2).
(l) et/4− e−t/4− sin(t)/2.
5. Write
x(t) = 6+ (6et−3 − 6)H(t− 3)+ (t− 4− 6et−3)H(t− 4)− (t− 4)H(t− 6).
7. F (s) = 2se
−3s − 2se
−4s.
9. f(t) = t3
6 e2t.
11. The equation can be written
x′′ + 4x = cos 2t− cos 2tH(t− 2π).
Transformed,
X(s) =s
(s2 + 4)2− e−2πs s
(s2 + 4)2.
Then
x(t) =1
4t sin(2t) +
1
4sin(2t)H(t− 2π)(2π − t).
13. The differential equation is q′′ + q = t − (t − 9)H(t − 9) with zero initialconditions. Taking transforms,
s2Q+Q =1
s2− 1
s2e−9s.
23
Hence,
Q(s) =1
s2(s2 + 1)− 1
s2(s2 + 1)e−9s.
Thereforeq(t) = t− sin t−H(t− 9)[t− 9− sin(t− 9)].
15. The differential equation is x′′ + π2x = π2 − π2H(t − 1). Taking thetransform and using the initial conditions,
(s2 − s+ π2)X =π2
s− π2
se−s.
Thus
X(s) =π2
s − π2
s e−s
s2 − s+ π2.
Use MATLAB to find x(t).
17. Write the square wave as
f(t) = 1− 2H(t− a) + 2H(t− 2a)− 2H(t− 3a) + 2H(t− 4a) + · · ·
= 1 + 2∞∑1
(−1)nH(t− na).
Take the transform to get
F (s) =1
s+ 2
∞∑1
(−1)n1
se−nas
=1
s
(1 + 2
∞∑1
(−e−as
)n)
=1
s
(−1 + 2
∞∑0
(−e−as
)n)
This series is a geometric series, and so it sums to
F (s) =1
s
(−1 +
2
1 + e−as
)=
1
s
1− e−as
1 + e−as.
Simplify to get 1s tanh(as/2).
19. Using the derivative formula,
L(f(t)) = F (s) = L(d
dt
∫ t
0
f(r)dr
)= sL
(∫ t
0
f(r)dr
)− 0.
24
21. Note that F (s) = arctan(a/s) and F ′(s) = −a/(s2 + a2). From Exercise20,
L−1 [F ′(s)] = −tf(t).
We want f(t), the inverse of F (s). We have
L−1[−a/(s2 + a2)
]= − sin at = −tf(t).
Thus f(t) = 1t sin at.
23. The solution involves changing the order of integration.∫ ∞
s
F (r)dr =
∫ ∞
s
(∫ ∞
0
f(t)e−rtdt
)dr
=
∫ ∞
0
(∫ ∞
s
f(t)e−rtdr
)dt
=
∫ ∞
0
f(t)
(∫ ∞
s
e−rtdr
)dt
=
∫ ∞
0
f(t)
[−1
te−rt
]∞s
dt
=
∫ ∞
0
f(t)
[1
te−st
]dt
=
∫ ∞
0
(f(t)
t
)e−stdt = L
[f(t)
t
].
Using this result we have
L[sinh t
t
]=
∫ ∞
s
1
r2 − 1dr
=
∫ ∞
s
(1/2
r − 1+
−1/2
r + 1
)dr
= −1
2lns− 1
s+ 1.
25. Taking the transform of the the two equations we get
sX − a = 2X − Y, sY = X.
Solving for X gives
X(s) =as
(s− 1)2= a
1
s− 1+ a
1
(s− 1)2.
Taking inverse transforms,
x(t) = aet + atet.
Then y(t) =∫x(t)dt.
25
� Sec. 3.3, page 162
1. (a) Use the sum formula for sine to get∫ t
0
cos τ sin(t− τ)dτ =
∫ t
0
cos τ(sin t cos τ − sin τ cos t)dτ.
Calculate the integrals in the standard way.
(b) ∫ t
0
e−2τe−3(t−τ)dτ = e−3t
∫ t
0
eτdτ = e−3t(et − 1).
(c) ∫ t
0
τ2(t− τ)dτ =1
12t4.
3. Transforming the equation gives
sX − x0 − aX = F (s).
Solve for X(s) to get
X(s) =x0s− a
+1
s− aF (s).
Therefore, x(t) = x0eat + eat ⋆ f(t), or
x(t) = x0eat +
∫ t
0
ea(t−τ)f(τ)dτ.
5. Use the substitution r = t− τ , dr = −dτ to get∫ t
0
x(τ)y(t− τ)dτ = −∫ 0
t
x(t− r)y(r)dr =
∫ t
0
y(r)x(t− r)dr.
7. Taking transforms,
s2 − 4X =1
s− 1
se−s.
Then
X(s) =1
s(s2 − 4)− 1
s(s2 − 4)e−s.
Thus
x(s) =e−2t + e2t
8− 1
4−H(t− 1)
(e−2(t−1) + e(2(t−1)
8− 1
4
).
26
9. The transformed equation is
s2X + 2sX + 2X =ω
s2 + ω2.
Thusx(t) = e−t sin t ⋆ sinωt.
11. Take transforms of the equation to get
s2 − sX = F (s) or X(s) =1
s(s− 1)F (s).
Taking inverse transforms,
x(t) = −(1− et) ⋆ f(t) = −∫ t
0
(1− eτ )f(t− τ)dτ.
13. Write
L−1
[1
(s2 + 1)2
]= L−1
[1
s2 + 1
1
s2 + 1
].
By convolution, the inverse transform is
sin t ⋆ sin t =
∫ t
0
sin τ sin(t− τ)dτ =1
2(sin t− t cos t) .
15. (a) Write the equation as
x(t) = t− t ⋆ x(t).
Then take transforms, using the convolution theorem, to get
X(s) =1
s2− 1
s2X(s).
Solving for X(s) and simplifying gives
X(s) =1
s2 + 1.
Thus, x(t) = sin t.
(b) x(t) = et/2.
(c) In the transform domain,
X(s) =1
s+ 1+
1
sX(s) or X(s) =
s
s2 − 1.
Thus x(t) = cosh t.
27
(d) As the problem stands, the solution is x(t) = 0. A better problem is
x(t) = −2 +
∫ t
0
cos(t− r)x(r)dr.
Then
X(s) = −2
s
s2 + 1
s2 − s+ 1.
Inverting,
x(t) = −4
3
√3et/2 sin(
√3 t/2)− 2.
17. The problem should read:
f(t) =1√π
∫ t
0
x(τ)√t− τ
dτ.
� Sec. 3.4, page 173
1. e−2.
3. Taking the transform of the equation and applying the initial conditionsgives
s2X −X = e−5s or X(s) =1
1− s2e−5s.
Inversion, using the switching theorem, gives x(t) = sinh(t− 5)H(t− 5).
5. f(t) = H(t− 2) + δ3(t).
7. Taking the transform of the equation and applying the initial conditionsgives
s2X +X − 1 = 3e−2πs or X(s) =1
s2 + 1+ 3
1
s2 + 1e−2πs.
Thenx(t) = sin t+ 3 sin(t− 2π)H(t− 2π).
9. Taking the transform of the differential equation and solving for X(s)gives
X(s) =1
2s2 + s+ 2e−5s,
which is the impulse response in the transform domain. The transferfunction isK(s) = 1/(2s2+s+2). In the time domain the impulse responseis k(t) ⋆ δ5(t) = H(t − 5)k(t − 5), where k(t) = L−1[K(s)]. We have, bycompleting the square in the denominator,
K(s) =1
2s2 + s+ 2
=1
2
√16/15
√15/16
(s+ 1/4)2 + (√15/16)2
.
28
The inverse of this expression is
k(t) =1
2
√16/15 e−t/4 sin(
√15/16 t).
This plots as a decaying oscillation.
29
Chapter 4 Exercises
� Sec. 4.1, page 190
1. (a) 19x
2 + 116y
2 = 1. An ellipse turning clockwise beginning at (0, 1).
(b) 19x
2 − y2 = 1. Right branch of a hyperbola.
(c) x = y2/4− 1. Parabola opening to the right.
(d) y = −2x. A straight line.
3. (a) Immediately x(t) = c1et. Substituting into the y equation and solving
(by an integrating factor) gives y(t) = −c1et + c2e2t. Thus,
x(t) = c1et
(1−1
)+ c2e
2t
(01
).
Or,
x(t) = c1
(et
−et)+ c2
(0e2t
).
(c) Note x′′ − x′ − 2x = 0. The eigenvalues are λ = 2, −1. Thus x(t) =c1e
2t+c2e−t. Substituting into the y equation gives y(t) = 1
2c1e2t−c2e−t.
Then
x(t) = c1
(e2t12e
2t
)+ c2
(e−t
−e−t
).
Problems (b) and (d) are similar.
� Sec. 4.2.1, page 198
1. For example,(1 32 4
)−1
=1
−2
(4 −3−2 1
)=
(−2 3/21 −1/2
).
3. (a) We have
det(A− λI) = det
(1− λ 32 4− λ
)= 0
givesλ2 − 5λ− 2 = 0.
The eigenvalues are λ± = 5/2±√33/2.
(b) Take each eigenvalue and substitute into (A − λI)x = 0. First, takeλ+. We have (
1− λ+ 32 4− λ+
)(x1x2
)=
(00
).
30
These two equations are the same (the determinant is zero), so solutionsare given by
(1− λ+)x1 + 3x2 = 0.
Taking x1 = 1 gives x2 = − 13 (1 − λ+). Therefore an eigenvector corre-
sponding to λ+ is (1
−13 (1− λ+)
)=
(1
12 +
√336
).
Make a similar calculation with λ−.
5. detA = 0 and therefore A−1 does not exist. All solutions of Ax = 0 aregiven by y = 2x which plots as a straight line of slope 1/2. Geometricallythis is the nullspace. Analytically, the nullspace is the set of all pointsx = α(1 2)T where α ∈ R.
7. By definition of the determinant on page 193,
det
(a b0 d
)= ad.
Similarly it holds for a zero in the upper right corner. The result holdsfor square matrices of all sizes.
� Sec. 4.2.2, page 201
1. (a) The coefficient matrices are:(−2 −3−1 4
).
(b) (0 −3−2 1
).
(c) (−2 01 0
).
3. (a) Setting both equations equal to zero gives the critical point x∗ = 14,y =∗= 28/3. Now use y = x − x∗ to obtain y′ = Ay, a homogeneousequation for y. See Equation (4.25).
(b) Setting both equations equal to zero gives gives the critical pointx∗ = −9/5, y =∗= 7/5. Proceed as in (a).
31
� Sec. 4.3, page 206
1. These 15 problems (a)–(o) are amenable to easy computation using soft-ware or an advanced calculator. By hand calculation, follow the proce-dures in Examples 4.27 and 4.28. Here, an additional example is workedout.
(b) Consider
A =
(1 14 1
).
The eigenvalues are found from det(A = λI) = λ2 − 2λ− 3 = (λ− 3)(λ+1) = 0. Therefore, λ = 3 and λ = −1 are the two eigenvalues. Now findassociated eigenvectors. Take λ = 3. Then the system (A − λI)x = 0takes the form (
1− λ 14 1− λ
)(x1x2
)=
(00
),
or, (−2 14 −2
)(x1x2
)=
(00
),
The two equations represented by this matrix set are the same. So −2x1+x+ 2 = 0, giving x1 = 1, x2 = 2. Therefore an eigenvector correspondingto λ = 3 is (1 2)T.
When λ = −1 the system takes the form(2 14 2
)(x1x2
)=
(00
),
Taking the first equation, 2x1 +x2 = 0, so x1 = 1, x2 = −2. Therefore aneigenvector corresponding to λ = −1 is (1 − 2)T.
3. Eigenvalues λ are solutions to det(A − λI) = 0. If λ = 0 then clearlydetA = 0. If detA = 0 then Ax = 0 has a nontrivial solution x. Thismeans λ = 0 is an eigenvalue since Ax = 0x must have a nontrivialsolution.
5. If Ax = λx then A2x = A(Ax) = A(λx) = λ(Ax) = λ2x. Then A3x =A(A2x) = A(λ2x) = λ2(Ax) = λ3x. By induction, Anx = λnx.
7. From Ax = λx we get(0 1−2 −3
)(1−2
)= λ
(1−2
).
Multiplying out and comparing components gives λ = −2.
32
� Sec. 4.4.1, page 217
1. (a) The general solution is
x(t) = c1
(−12
)e−t + c2
(12
)e−2t.
(0, 0) is a stable node.
(b) The general solution is
x(t) = c1
(−12
)e−t + c2
(12
)e2t.
(0, 0) is a unstable saddle.
(c) The general solution is
x(t) = c1
(−12
)et + c2
(12
)e2t.
(0, 0) is a unstable node.
(d) The general solution is
x(t) = c1
(21
)e−3t + c2
(−10
)e2t.
(0, 0) is a unstable saddle.
3. (a) The eigenvalues are −1, −3, and therefore (0, 0) is a stable node.
(b) The determinant is −4 and so (0, 0) is a saddle point.
(c) detA = 1 and trA = −3. Therefore (trA)2 − 4 detA = 5 > 0, and(0, 0) is a stable node.
(d) detA = 6 and trA = 4. Therefore (trA)2 − 4 detA = −8 < 0, and sothe eigenvalues are complex. The origin is an unstable spiral.
5. The general solution is
x(t) = c1
(13
)e2t + c2
(11
)e4t.
The initial condition is
x(0) = c1
(13
)+ c2
(11
)=
(2−1
).
This gives c1 + c2 = 2, 3c1 + c2 = −1. Solving gives c1 = −3/2, c2 = 7/2.
33
� Sec. 4.4.2, page 220
1. (a) One complex fundamental solution is
x(t) =
(1i
)eit
=
[(10
)+ i
(01
)](cos t+ i sin t)
=
(cos t− sin t
)+ i
(sin tcos t
).
The real and imaginary parts of this expression are two real fundamentalsolutions. Then the general solution to the system is
x(t) = c1
(cos t− sin t
)+ c2
(sin tcos t
).
(b) As in part (a), one complex solution is
x(t) =
(1
1 + i
)e(1+2i)t
=
[(11
)+ i
(01
)](cos 2t+ i sin 2t)et.
Separate real and imaginary parts to find two real fundamental solutionsand thus the general solution.
3. The system isx′ = 3x+ 4y, y′ = x− 3y.
The trace of A is 0 and the determinant is −13 < 0. Thus (0, 0) is a saddlepoint. The x and y nullclines are y = −3x/4, y = x/3, respectively.Note x′ > 0 when y > −3x/4 and y′ > 0 when y < x/3. The eigenvaluesare λ = ±3.6056 with corresponding eigenvectors [0.9887 0.1497]T and[0.5180 0.8554]T, respectively. The eigenvectors define the directions ofthe separatrices.
� Sec. 4.4.3, page 224
1. (a) The eigenvalues are obviously repeated, −3 and −3. There is only oneeigenvector, v = [1 0]T and the system is deficient. One solution is
x1(t) =
(10
)e−3t.
We must find a generalized eigenvectorw satisfying (4.39): (A+3I)w = v,or simply
w1 = 0, w2 = 1.
34
Therefore another independent solution of the equation is
x2(t) = (tv +w)e−3t =
[t
(10
)+
(01
)]e−3t.
(b) The eigenvalues are 2, 2 with only one eigenvector v = [−1 1]T. Thematrix is deficient. One solution is
x1(t) =
(−11
)e2t.
To find a generalized eigenvector w we solve (A− 2)w = v. We have(−1 −11 1
)(w1
w2
)=
(−11
).
A solution is
w =
(10
).
So the general solution to the system is
x(t) = c1
(−11
)e2t + c2(tv +w)e2t.
� Sec. 4.5, page 235
1. (a) The eigenpairs are 1, [2 1]T and 3, [1 1]T.
(b) The eigenpairs are 0, [−3 2]T and 8, [1 2]T.
(c) The eigenpairs are ±2i, [0.9428 0.2357± 0.2357i]T.
3. The general solution is
x(t) = c1
(12
)e−6t + c2
(1−5
)e−t.
The linear orbits are y = 2x and y = −5x. The origin is a stable node,and the other orbits enter the origin tangent to the direction y = 2x ast → ∞. As t → −∞ all orbits become parallel to y = −5x. The orbitbeginning at the point (0, 1) satisfies
x(0) = c1
(12
)+ c2
(1−5
)=
(01
),
which gives c1 = −c2 = 17 .
5. The eigenvalues λ are 1, 1 with a single eigenvector v = [1 − 1]T. Thesystem is deficient. We must find a generalized eigenvector w satisfying(A− λI)w = v. This system is(
1 1−1 −1
)(w1
w2
)=
(1−1
),
35
which has a solution w1 = 1, w2 = 0. Therefore the general solution is
x(t) = c1
(1−1
)et + c2
(t+ 1−t
)et.
To solve the initial value problem we set t = 0 to get
c1 = 1, c2 = 0.
Therefore
x(t) =
(et
−et).
7. When α = 12 , (0, 0) is a stable node. When α = 2, (0, 0) is a saddle point.
Therefore, as α varies from 1/2 to 2 the system undergoes a bifurcationfrom a stable node to a saddle. In general, for arbitrary α in this range,the trace is −2 and the determinant is 1−α. So the bifurcation occurs atα = 1, where the determinant changes sign. When α = 1 the eigenvaluesare 2 and 0 and the matrix has determinant zero; thus there is a liney = −x of nonisolated critical points.
9. We have tr(A) = 4 > 0 and detA = 3 − a. If a > 3then (0, 0) is anunstable saddle point. If a = 3 then (0, 0) is a non-isolated critical pointand there is a line of critical points along y = −x. These are all unstableand orbits are straight lines exiting the critical points. If a < 3 then thenwe check the discriminant tr(A)2−4 detA = 4(1+a). If −1 ≤ a < 3, then(0, 0) is an unstable node, and if a < −1 then (0, 0) is an unstable spiral.
11. (a) trA = 3, detA = −4; saddle point.
(b) The eigenvalues are −3, −3; stable node.
(c) trA = 5, detA = −6; saddle point.
(d) trA = −15, detA = 44; (trA)2 − 4 detA > 0; stable point.
(e) The eigenvalues are 2, 2; unstable node.
(f) trA = 2, detA = 5; (trA)2 − 4 detA = −16 < 0; unstable spiral.
(g) trA = 6, detA = 9; (trA)2 − 4 detA = 0; unstable node.
(h) trA = 0, detA = 9; center.
13. With the parameters as given the coefficient matrix is
A =
(−4 1
44 −4
).
The eigenvalues are −3 and −5, so the origin is an asymptotically stablenode. The general solution is
x(t) = c1
(0.24250.9701
)e−3t + c2
(−0.24250.9701
)e−5t.
36
The initial conditions are C1(0) = 0 and C2(0) = 0.3. Apply these to findc1 and c2 in the general solution by solving
c1 − c2 = 0, c1 + c2 = 0.3092.
Adding, we get c1 = 0.1546; subtracting we get c2 = 0.1546.
15. (c) Substitute I3 = I1 + I2 into part (a) to obtain a system for I1 and I2.In matrix form,
L
(1 01 1
)(I ′1I ′2
)=
(−R1 R2
−R3 −R2 −R3
)(I1I2
)+
(E(t)0
).
Multiply through by the inverse of the matrix on the left side gives(I ′1I ′2
)=
1
L
(−R1 R2
R1 −R3 −2R2 −R3
)(I1I2
)+
(1LE(t)0
).
Then the trace of the matrix on the right is obviously negative and thedeterminant is
det =1
L(R1R2 +R1R3 +R2R3) > 0.
Therefore the homogeneous system has an asymptotically stable criticalpoint at the origin.
17. Substitute the proposed solution into the equation to get
eσtσv = rλveσte−σT ,
where we used the fact that Av = λv. Making cancelations, we get
σ = rλe−σT .
Assume T and r are positive and take the case that λ > 0. Proceed geo-metrically and plot the left side and right side of the equation as functionsof σ. Then, if σ > 0 the right side is a decaying exponential and the leftside is a straight line; there will always be a unique positive solution. Ifσ < 0 then there are no intersections.
� Sec. 4.6, page 243
1. The first part is straightforward: compute Φ′(t) and then AΦ and showthat the entries match. For the second part,
x(t) = c1
(ϕ1ϕ2
)+c2
(ψ1
ψ2
)=
(c1ϕ1 + c2ψ1
c1ϕ2 + c2ψ2
)=
(ϕ1 ψ1
ϕ2 ψ2
)(c1c2
).
37
3. Use Equation (4.45). We have
Φ(t) =
((1 + t)e2t −te2tte2t (1− t)e2t
),
and so the inverse is
Φ−1(t) =
((1− t)e−2t te−2t
−te−2t (1 + t)e−2t
).
(The determinant of Φ(t) = e4t.) Therefore, by (4.45), the general solutionis
x(t) =
((1 + t)e2t −te2tte2t (1− t)e2t
)×[(
k1k2
)+
∫ t
0
((1− s)e−2s te−2s
−te−2s (1 + s)e−2s
)(0s
)ds
].
Clearly,
x(0) = Φ(0)k =
(1 00 1
)(k1k2
)=
(12
).
so k1 = 1, k2 = 2.
5. The fundamental matrix is
Φ(t) =
(cos t sin t− sin t cos t
).
Now use undetermined coefficients to determine a particular solution.Take
xp(t) =
(A cosωt+B sinωtC cosωt+D sinω
).
Substitute into the nonhomogeneous equation, noting that ω ̸= ±1, givesB = C = 0 and
A =1
1− ω2, D = − ω
1− ω2.
Therefore, a particular solution is
xp(t) =
( 11−ω2 cosωt
− ω1−ω2 sinω
).
The general solution is x(t) = Φ(t)c+ xp(t).
7. The eigenvalues are 4, 4 with one eigenvector v = [−1 1]T. The system isdeficient and we need a generalized eigenvector w satisfying (A−λI)w =v. Easily a solution is w = [1/6 1/6]T. The fundamental matrix is there-fore
Φ(t) =
(−e4t e4t
(−t+ 1
6
)e4t e4t
(t+ 1
6
) ).
38
9. The equations are
x′ = −r1x− r3x+ r2y +D, y′ = r1x− r2y.
The longtime behavior is the equilibrium conditions x′ = y′ = 0. Solvingthese two equation for x and y gives
x =r1 +D
r1 + r2, y =
r1r2.
39
Chapter 5 Exercises
� Sec. 5.1, page 257
3. There are no critical points because x′ is never zero. There is one nullcline,y′ = 2x = 0, which is the y axis; there the vector field is horizontal. Tofind the integral curves, divide the two equations to get
dy
dx= −2xy.
Separating variables and integrating gives bell-shaped integral curves
y = Ce−x2
.
5. The critical points are (±1,−1). The nullclines are y = −1 (vector fieldvertical) and y = −x2 (vector field horizontal). Using the Jacobian,J(1,−1) has negative determinant and thus (1,−1) is a saddle point;J(−1,−1) has trace = 1 and det = 2 and therefore (−1,−1) is an un-stable spiral. To find the vector field in regions between the nullclines,note x′ > 0 when y > −1 and y′ > 0 when y > −x2, and so on.
7. There is a line of nonisolated critical points y = x. A nullcline is y = −xwhere the vector field is vertical. We can find equations of orbits bydividing the equations to get
dx
dy= x+ y.
This is a linear equation for x = x(y) and an integrating factor is e−y.
Solving gives integral curves
x = −1− y + Cey.
When x = 1/4, y = 0, we get C = −3/4. For any fixed value of C, thecorresponding integral curve cannot cross the line y = x; it either entersor exits that line. See also Exercise 9.
9. The entire x axis is a nonisolated set of critical points. The line x = −2 isa nullcline where the vector field is horizontal; above the x axis it pointsE, and below the x axis it points W. The integral curves are found bydividing the two equations and integrating: y = 1
2x2 + 2x + C, y ̸= 0.
A parabola cannot cross the x axis. Two typical orbits are shown in theFigure 2.
40
x
y
criticalpoints
Figure 2: Exercise 9. Parabolic orbits entirely above the x axis move left toright. The other parabolas are broken at the x axis as shown with their givedirections.
11. Six critical points:
(0, 0), (0, 3/2), (0,−3/2), (2, 0),
(1± 1
2√2, 1∓ 1
2√2
).
The x nullclines are x = 0 and y = 2 − x; the y nullclines are y = 0 andthe circle x2 + y2 = 9/4. To test stability, compute the Jacobian matrix
J(x, y) =
(2− 2x− y −x
−2xy −3y2 − x2 + 94
)at the six critical points. For example,
J(0, 0) =
(2 00 9
4
),
so the matrix J(0, 0) has positive eigenvalues; thus (0, 0) is an unstablenode.
13. The critical points are (0, 0) and (2, 4). Using the Jacobian we see (2, 4)is a saddle point, but the Jacobian gives no information at (0, 0) becausedetJ(0, 0) = 0. However, x = 0 (y axis) is a nullcline and represent twoopposing orbits, both approaching the origin. The vector field has a stablenodal structure at (0, 0) in the right half-plane. In the left half-plane thelocal behavior near (0, 0) is that of a saddle point.
15. The only critical point is (0, 0). For the linearized system det J(0, 0) = 1and trJ(0, 0) = 1, which gives no information. However, on any family ofcircles x2 + y2 = C we have
d
dt(x2 + y2) = 2xx′ + 2yy′ = −2y4 < 0.
41
Figure 3: Exercise 3. Orbit with initial condition x = 1, x′ = y = 0.
Thus, the orbits must approach the origin and hence (0, 0) is asymptoti-cally stable.
� Sec. 5.2, page 269
1. The only critical point is (0,0), and the Jacobian matrix is
J(0, 0) =
(0 11 0
).
Thus the origin is a saddle point because detJ(0, 0) < 0.
3. For x′′ = −x2 the force is F (x) = −x2 and the potential energy V (x) =∫x2dx = 1
3x3. The conservation law is
1
2y2 +
1
3x3 = E =
1
3.
Thus
y = ±√
2
3
√1− x3.
The orbit beginning at (1, 0), shown in Figure 3, is the branch in the lowerhalf-plane y < 0.
5. The potential is V (x) = 12x
4 and the conservation law is x4 + y2 = C,where C = 2E. This plots a clockwise, closed loop around the origin.When x(0) = x0, y(0) = 0, we get C = x40. So the conservation law canbe written
y =dx
dt= ±
√x40 − x4.
Separating variable and integrating from 0 to x0 (one-fourth of a periodT ) gives ∫ x0
0
dx√x40 − x4
=T
4,
42
where we have taken the + sign since the velocity is positive in y > 0.In the integral on the left make the substitution x = x0r, dx = x0dr toobtain, after simplification,
1
x0
∫ 1
0
dr√1− r4
=T
4.
7. The potential energy is
V (x) =
∫gR2
(x+R)2dx = − gR2
x+R+ C.
When x = 0 we get C = gR. Therefore the conservation law is
1
2y2 − gR2
x+R+ gR = E =
1
2y20 .
Rearranging and simplifying gives the orbits depending on the initial ve-locity y0:
y = ±
√y20 − 2gR
(1− R
x+R
).
To easily plot phase paths we factor out 2gR to get
y = ±√2gR
√α− x
x+Rwhere α ≡ y20
2gR.
Plot the right side of this equation for three different values of α, sayα1 < 1, α2 = 1, α3 > 1. The result is shown in Figure 4. For α < 1the mass velocity becomes zero at a finite value of x, and it then turnsaround back toward the earth with a negative velocity. For α ≥ 1 the massmaintains a nonzero, positive velocity and escapes the earth’s gravitationalfield. The value α = 1 is therefore the smallest value in which the particleescapes; thus, y0 =
√2gR is called the escape velocity.
The radius of the earth is R = 6.731 million meters, and g = 9.8 metersper second. Therefore the escape velocity is 11.2 km/s or 25, 000 mph.
9. Refer to Figure 5.9 in the text. The force on the mass is =mg, downward.We decompose the force into two components, one normal (perpendicular)to the arc of the path, and one tangent to the arc. The normal componentcauses no motion; the tangential force is−mg sin θ. Therefore the equationof motion, mass times acceleration equals force, is
ms′′ = mlθ′′ = −mg sin θ.
Here we used s = lθ, where s is arclength. The result follows.
43
x
x
y
y
x/(x+R)
a
a
a
a
a
a
1
1
2
2
3
3orbits
Figure 4: Exercise 7. Orbits for α1 < 1, α2 = 1, α3 > 1, where α ≡ y20
2gR .The value α = 1 is the smallest value for which the mass escapes the earth’sgravitational field.
11. By Taylor’s formula,
sin θ = θ − 1
3θ3 +
1
5!θ5 − · · · .
For small θ we take only the first approximation sin θ ≈ θ. Thus theequation of motions (see Exercise 9) becomes
θ′′ = −glθ.
We recognize this as an oscillator equation (Chapter 2). We try a solutionθ(t) = A cosωt and substitute to get ω =
√g/l. Clearly, A = θ(0).
The period is T = 2π/√g/l. For the crane problem, we have θ(t) =
A cos(√g/20 t), so the ball hits the building when A cos
√g/20 t∗ = 0, or
when√g/20 t∗ = π/2. Then
t∗ =π
2
√20/g = 2.244 seconds.
13. We have
d
dtV (x) = 2xx′ + 2yy′ = 2xy + 2y(−x− y3) = −2y4 < 0.
Therefore V is decreasing in time.
44
15. Refer to Exercise 9. Including the damping force −kθ′, we get from New-ton’s second law
mlθ′′ = −mg sin θ − kθ′.
Thus
θ′′ = −glsin θ − k
ml.
We can write this as a system
θ′ = ω, ω′ = −glsin θ − k
ml.
Therefore, if E = 12ω
2 + (g/l)(1− cos θ), then
E′ = ωω′ +g
lsin(θ)θ′,
or
E′ = ω
(g
lsin θ − k
mlω
)+ ω − g
lsin θ = − k
mlω2 < 0.
E is the total kinetic energy plus the potential energy due to conservativeforces. The last equation states that this energy dissipates at the ratekmlω
2.
� Sec. 5.3.2, page 282
1. The critical points are(h
r, 0
), P =
(m
b,rm− bh
am
).
We assume rm > bh, which is valid if the harvesting rate is small. Noticethat the coexistent state, given by the second critical point, has the sameprey value as the Lotka–Volterra model, but the predator equilibrium issmaller. Therefore, seemingly ironic, it is the predator population thatdecreases when the prey is harvested. From the Jacobian it easily followsthat (h/r, 0) is a saddle point. At the second critical point the Jacobianis
J(P ) =
(bh/m −am/b
b(rm− hb)/am 0
).
The trace is positive and the determinant is positive. Therefore the equi-librium P is unstable, either a node or a spiral. Aspects of the phaseplane are shown in Figure 5. This appears not to be a reasonable modelof harvesting.
3. Note that x′ = 0 wheny =
rx
a(x− k)
45
m/bh/r
r/a
x
y
P
Figure 5: The phase diagram for Exercise 1 showing the nullclines and criticalpoints P (unstable) and (h/r) (saddle point). A sample orbit is shown.
and y′ = 0 when y = 0 or y = k +m/b. These nullclines are plotted inFigure 6. The critical points are (0, 0), a saddle point, and
P =
(k +
m
b,r
a+
bk
am
).
One checks the Jacobian matrix to see that, evaluated at this critical point,the trace is negative and determinant is positive. So P is asymptoticallystable. In conclusion, when a refuge is available, the coexistent statebecomes asymptotically stable rather than a center for the case of norefuge.
5. The critical points are (0, 0), (m/b, k/a). The Jacobian matrix is
J(x, y) =
(−k + ay ax
by −m+ bx
).
The Jacobian at the origin has eigenvalues −k, −m and so the origin is astable node. The nonzero critical point has eigenvalues ±
√mk, and is a
saddle point. See Figure 7.
7. The critical points are (0, 0), (0, 32), (28, 0), (12, 8). The Jacobian matrixis
J(x, y) =
(14− x− y −x
−y 16− x− y
).
J(0,0) has eigenvalues 14, 16 and gives an unstable node; J(0,32) haseigenvalues -18,-16 and gives a stable node; J(28,0) has eigenvalues -14 -12 and gives a stable node. It is easily checked that (12,8) is a saddle point
46
(0,0)
r/a
k+m/bk x
y
.
asymptote
asymptote
P
Figure 6: The phase diagram for Exercise 3 showing the direction field, null-clines, and critical points P and (0, 0). Also shown are the vertical asymptotex = k and the horizontal asymptote for the x nullcline.
Figure 7: The form of the phase plane diagram for Exercise 5. Any initialcondition below the darker (NW to SE) separatrices leads to extinct populations,(0, 0). There can be no coexistent states. If the initial condition were on a darkerseparatrix leading into the equlilbrium ((m/b, k/a), small perturbations, whichare always present, would knock the path off the separatrix and the populationwould go extinct or blow up.
47
(28,0)
(12,0)
(0,0)
Figure 8: The form of the phase plane diagram for Exercise 7.
either by the Jacobian or the form of the direction field and nullclines. SeeFigure 8.
9. The x nullclines are x = 0 and x = ay/(1 + y). The y nullcline is
y =(b− y)(1 + y)
2a.
Because we have simple formulas for the nullclines for x in terms of y, itis reasonable to switch the x and y axes in our plots. (One could alsorotate the axes to obtain the usual plots.) Figure 9 shows the nullclines,direction field, and critical points (0, b) and P . To find P add the twodifferential equations to get x = b − y, a relation that also must hold atthe critical point. Setting
ay
1 + y= b− y
and solving for y gives the y coordinate of P . However, we do not need tofind the critical point P to determine stability. First, the Jacobian evalu-ated at (0, b) is easily and has eigenvalues ab/(1 + b) and −1. Therefore,(0, b) is a saddle point. The Jacobian matrix for P has sign structure
J =
(− +− −
).
Hence det J > 0 and tr J < 0, P is asymptotically stable.
11. Note a typo. The second equation should read
dG
dt= c2G(1−G)− c1TG− d2G.
13. The x nullclines are x = 0 and y = 4 − x; the y nullclines are y = 0 andy = (2−bx)/a. To have a coexistent state the two nullclines y = 4−x and
48
y
x
(0,b)
P
Figure 9: Phase diagram for Exercise 9. Note that the axes are switched forease of plotting.
4 4
44
2/a
2/a
2/b 2/b
case 1 case 2
Figure 10: The form of the phase plane diagram for Exercise 13 in two cases.The figure on the right represents the case when a coexistent state exists.
49
S
I
N
N
r/a
(S(0),I(0))
Figure 11: Exercise 1. When r/a > N there is no epidemic and the diseaseimmediately dies out. The phase diagram shows the direction field computedfrom the differential equations and an orbit beginning at (S(0), I(0)) on the lineS + I = N .
y = (2− bx)/a must intersect in the first quadrant. This will occur in twocases: Case 1: 2/a > 4 and 2/b < 4, and Case 2: 2/a < 4 and 2/b > 4. Seefigure 10, which shows both cases. It is clear from the direction field thatin Case 1 the interior critical point has a saddle-like structure, whereas inCase 2 it is an asymptotically stable node. So the competing species cancoexist in Case 2. Note that 2/a is the carrying capacity of species y andb is its competition factor.
� Sec. 5.3.3, page 292
1. The phase diagram in the case r/a > N is shown if Figure 11,
3. Here, N = 200, I0 = 20 and S∗ = 100. It takes 3 days to recover, sor = 1
3 . To compute the infection constant a use the fact that S∗ satisfies
−S∗ +N +r
alnS∗
S0= 0, S0 = 180.
Substitute the values into this equation and solve for a to get a = 0.00196.The initial infection rate is I ′ = aS(0)I(0) = 0.00196(180)(20) ≈ 7. Soabout 7 individuals get the disease per day.
5. For an SI disease the compartmental diagram is S I, with the S → Irate aSI and the I → S rate rI. The equations are
S′ = −aS + rI, I ′ = aSI − rI, S + I = N.
Substitute S = N − I into the I ′ equation to get, after simplification,I ′ = aI(N − I) − rI = aI(N − I − r/a). This has stable equilibria atI = N − r/a, which is the limiting number of infections. Therefore, Iincreases to this value. Notice that the equation for I is similar to thelogistic equation. Here we assume N > r/a. If N < r/a, the infection diesout.
50
AIDS
x y
m m c
axy/(x+y)
Figure 12: Exercise 9. HIV compartmental diagram. µ is the per capita mor-tality rate and c is the per capita rate HIV infectives get AIDS.
7. The governing equations are
S′ = −aSI + µ(N − S − I), I ′ = aSI − rI,
where we used R = N −S− I. There are at most two equilibria, at (N, 0)and (
r
a, µN − r/a
µ+ r
).
The first is a state with no infectives, and the second is an endemic state.The Jacobian matrix is
J =
(−aI − µ −aS − µaI aS − r
).
At the critical point (N, 0) the eigenvalues are −µ and aN−r. If the latteris positive, which we assume, then (N, 0) is a saddle point. At the secondequilibrium point a straightforward calculation shows that the trace isnegative and the determinant is positive. Therefore it is asymptoticallystable. This represents an endemic state where the disease has becomeestablished in the population. We leave the the sketch of the phase planeto the reader. Use the nullclines and direction field to obtain the diagram.
9. The equations are
x′ = b− µx− axy
x+ y,
y′ = axy
x+ y− µy − cy.
A compartmental diagram is in Figure 12.
13. The diagram isS → E → I,
where the first rate is aSI and the second is kE. The equations are
S′ = −aSI, E′ = aSI − kE, I ′ = kE.
51
S
I
N
N
Figure 13: Phase diagram for Exercise 13. All orbits approach the state (0, N)where everyone is infected.
Using E = N − S − I the S and I equations become:
S′ = −aSI, I ′ = k(N − S − I).
The critical points are (N, 0) and (0, N). The Jacobian shows easily thatdetJ(N, 0) = −akN < 0, so (N, 0) is a saddle point. And J(0, N) haseigenvalues −aN and −k, so (0, N) is a stable node. A phase diagram isshown in Figure 13.
� Sec. 5.3.4, page 300
1. For part (a) make the substitution t = τ/µ in the differential equations asindicated.
(b) In the first differential equation neglect the εh term to get
h′ = λm(1− h), m′ = ηh(1−m)−m.
Here, ‘prime’ denotes d/dτ . The nullclines are h′ = 0 when m = 0 orh = 1, and m′ = 0 when m = ηh/(ηh+ 1). There are two critical points,(0, 0) and (1, η/(η + 1)). The Jacobian is
J(h,m) =
(λm λ(1− h)
η(1−m) −ηh− 1
).
It is easy to check det J(0, 0) = −ηλ, so (0, 0) is a saddle point. Also,
J(h,m) =
(−ηλη+1 0η
η+1 −η − 1
).
Both eigenvalues (diagonal elements) are negative and therefore (1, η/(η+1)) is an asymptotically stable node. It represents the case that all hu-mans are infected, but only a fraction of mosquitos are infected. A phasediagram is shown in Figure 14.
52
h
m
1(0,0)
1
P
Figure 14: Phase plane for Exercise 1b. The dashed line is the m nullcline. Alloribits approach the critical point P= (1, η/(η + 1)).
� Sec. 5.4, page 306
1. Transform to polar coordinates as in Example 5.13.
3. Note fx+ gy = (1+3x2)+5y4 > 0, and so there can be no periodic orbitsby Dulac’s Theorem.
5. Note that
(βf)x + (βg)y =
(1
xyx(P − ax+ by)
)x
+
(1
xyy(Q− cy + dx)
)y
= −ay− c
x< 0
in x, y > 0, because a, c > 0. Therefore there are no periodic orbits.
7. Take
F (x, y) =x3 + y3
3xy
and show that ddtF (x(t), y(t)) = C using the chain rule.
9. (a) If f = Hy and g = −Hx then fx = Hyx and gy = −Hxy. By equalityof mixed partials we have fx = −gy, or fx + gy = 0.
(b) By the chain rule, ddtH(x, y) = Hxx
′+Hyy′ = −gf+fg = 0. Therefore
H is constant when the differential equations hold.
(c) Note that the Jacobian matrix is at an equilibrium point P is
J(x, y) =
(fx fygx gy
).
If P is a source or sink, then the trace of the matrix would be nonzero.But the trace is fx + gy = 0, contradicting this fact.
53
(d) x′′ = F (x) is equivalent to the system x′ = y, y′ = F (x). Notef(x, y) = y and g(x, y) = F (x). Hence, fx + gy = 0, and the system isHamiltonian. To find H we solve
Hx = −F (x), Hy = y.
Integrate the second equation with respect to y to get H(x, y) = 12y
2 +ϕ(x), where ϕ is an arbitrary function of x. Hence Hx = ϕ′(x). But fromthe other equation, Hx = −F (x). Thus, ϕ′(x) = −F (x). Integrating givesϕ(x) = −
∫F (x)dx + C. But −intF (x)dx = V (x), the potential energy.
Hence
H(x, y) =1
2y2 + V (x) = constant.
The total energy is conserved.
(e) x′ = y, y′ = x − x2. From part (a) the system is Hamiltonian, andfrom part (d)
1
2y2 − 1
2x2 +
1
3x3 = C.
11. (a) If there is a G such that f = Gx and g = Gy, then fy − gx = Gxy −Gyx = 0. So the scalar curl is zero. To prove the converse, that thereexists a G, we refer the reader to any multivariable calculus text book.
(b) By the chain rule, ddtG(x, y) = Gxx
′ + Gyy′ = f2 + g2 > 0. It fol-
lows that the scalar function G(x(t), y(t)) in increasing on every orbit.Therefore, there can be no periodic orbits and thus no centers.
(c) Part (b) shows a critical point cannot be a spiral point. The Jacobianat a critical point is
J(x, y) =
(fx fygx gy
)=
(Gxx Gxy
Gyx Gyy
).
The eigenvalues are
λ =1
2
(trJ ±
√(trJ)2 − 4 det J
).
We show that the discriminant is positive; so there are no complex eigen-values and thus no spiral points.
(trJ)2 − 4 detJ = (Gxx +Gyy)2 − 4(GxxGyy −G2
xy)
= (Gxx −Gyy)2 + 4G2
xy > 0.
(d) Note that fy − gx = 0, so the system is a gradient system. To find Gnote
f = Gx = 9x2 − 10xy2, g = Gy = 2y − 10x2y.
Integrate the first equation with respect to x to get
G(x, y) = 3x3 − 5x2y2 + ϕ(y).
54
Then finding Gx and comparing to the expression above gives ϕ(y) = 13y
3.Thus,
G(x, y) = 3x3 − 5x2y2 +1
3y3.
(e) Use the method in part (d) to find G(x, y) = x sin y + C.
13. (a) Factor, and write r′ = r(r − 2)r − 1), and θ′ = 1. Thus θ(t) = t+ C,and θ winds counterclockwise. The periodic orbits are then the clockwisecircles r = 1 and r = 2. Next note r′ > 0 for r > 2 and r < 1, whiler′ < 0 for 1 < r < 2. So the orbits with r(0) < 1 spiral out the originapproaching 1. Orbits starting with 1 < r(0) < 2 spiral into r = 1 ast → +∞ and into r = 2 as t → −∞. For r(0) > 2 the orbits spiraloutward to infinity as t→ +∞ and toward r = 2 as t→ −∞.
(b) Use the formulas in Example 5.13.
� Sec. 5.5, page 311
1. The matrix for this linear system is
J =
(ε −11 ε
).
Note that (trJ)2−4 det J = −4, so the eigenvalues are always complex, andthey are given by λ(ε) = ε± i. As ε varies from −0.2 to 0.2 the eigenvaluesmove in the upper complex plane along the straight line connecting−0.2+ito 0.2+ i, and similarly in the lower complex plane (complex conjugates).In this range the origin is a stable spiral (ε < 0), a center (ε = 0), andthen a stable spiral (ε > 0).
3. Write the equations as x′ = x(y − 1), y′ = y(4 − 4y/K − 2x). The xnullclines are x = 0, y = 1, and the y nullclines are y = 0, y = K(1−x/2).The critical points are (0, 0), (0,K); if K > 1 there is a third critical point
P =
(2K − 2
K, 1
), K > 1.
If K < 1 then this critical point is not present because x, y ≥ 0.
The Jacobian matrix is
J =
(y − 1 x
−2y 4− 8yK − 2x
).
The matrix J(0, 0) has eigenvalues−1, 4, so (0, 0) is a saddle point; J(0,K)has eigenvalues K−1 and −2. Therefore, if K > 1 then (0,K) is a saddle,and if K < 1 it is a stable node. In the special case K = 1 there are justtwo critical points and (0, 1) = (0,K) has eigenvalues 0,−2.
55
2(0,0)
(0,K)
1
x
y
Figure 15: Exercise 3. Phase plane in the case K > 1/2 +√5/2 when the
critical point P is a stable spiral. For 1 < K < 1/2 +√5/2 the point P is an
asymptotically stable node; the phase plane is similar but curves enter P rapidlywithout oscillation.
In the case of the critical point P , we have, with K > 1,
J(P ) =
(0 2K−2
K−2 − 4
K
).
The trace is negative and determinant positive; therefore P is asymptoti-cally stable. The discriminant is
tr2 − 4 det =16
K2
(1 +K −K2
).
This changes sign at 1/2 +√5/2. For 1 < K < 1/2 +
√5/2 the critical
point is a node, and for K > 1/2 +√5/2 the critical point is a spiral.
The phase diagrams are shown in the Figure 15. We leave the special caseK = 1 to the reader.
5. The quantities are: P is plant biomass and H is herbivores. ϕ is theplant biomass production (plant biomass/time), a is the death rate ofplants (1/time), b is the consumption rate (1/(herbivores· time)), c is themortality rate of herbivores (1/time), and ε is the yield (herbivores/plantbiomass). The nullclines are P ′ = 0 when H = (ϕ− aP )/bP and H ′ = 0when H = 0 or P = c/εb. The critical points are(
ϕ
a, 0
), Q =
(c
εb,εbϕ− ac
bc
).
The second does not exist unless εbϕ− ac ≥ 0, that is, the primary plantproduction is large enough. In this case the phase diagram is shown in
56
c/be c/bef/a f/a
H H
PPnode saddle
Q
Figure 16: Exercise 5. The left figure is low primary production ϕ, and on theright high primary production. The direction fields are shown. The orbits areleft to the reader.
Figure 16 (right panel). It is straight forward to draw in the directionfield. The Jacobian matrix is
J(P,H) =
(−a− bH −bPεbH −c+ εbP
).
In the case εbϕ− ac ≥ 0 we find that det J(ϕ/a, 0) = −εbϕ+ ac < 0, andtherefore (ϕ/a, 0) is a saddle point. At the critical point Q we find
trJ(Q) = −εϕc< 0, det J(Q) = εbϕ− ac > 0.
Thus Q is asymptotically and there is a coexistent state.
When there is only a single critical point (ϕ/a, 0), it is a stable node andthe herbivores die out. In summary, if ϕ is small the herbivores die outuntil it reaches the value ϕ = ac/εb. At that time the zero population ofherbivores becomes unstable and a new stable coexistent state is born.
7. (a) Take h = 0, a, b > 0, to get x′ = x(1− ax− y), y′ = y(b− y − x). Thenullclines are
x′ = 0 on x = 0, y = 1− ax; y′ = 0 on y = 0, y = b− x.
There are always the the critical points (0, 0), (0, b), (1/a, 0). If b < 1,a > 1, and b > 1/a, then there is a fourth critical point at
P =
(1− b
a− 1,ab− 1
a− 1
).
In this case the Jacobian matrix is
J(x, y) =
(1− 2ax− y −x
−y b− x− 2y
).
57
1
(0,b)
(0,0) b
P
(1/a,0)saddle
saddle stable node
unstable node
x
y
Figure 17: Exercise 7a. The critical points, nullclines, and direction field in thecase b > 1/a, b < 1, and a > 1.
A phase diagram showing the critical points and nullclines is shown inFigure 17. It is straightforward to classify the critical points by substitut-ing into the Jacobian matrix and checking the eigenvalues. We see (0, 0)is an unstable node, (0, b) and (1/a, 0) are saddles, and from the form ofthe direction field, P is clearly a stable node. We leave it to the reader todraw in the orbits.
We leave the other cases (b < 1/a, b > 1) to the reader. In these cases thecritical point P disappears. Parts (b) and (c) are involved and are left asexercises.
9. Note x′ = 0 when y = −(1− x)(2− x) and y′ = 0 when y = ax2. So thenullclines are parabolas; the x nullcline is fixed and the y nullcline changeswith the positive parameter a. The three cases are shown in Figure 18.The x coordinate of critical points are roots of (a+ 1)x2 − 3x+ 2 = 0, or
x =1
2(a+ 1)(3±
√1− 8a).
So, there are no critical points for a > 1/8, one critical point for a = 1/8,and two critical points for a < 1/8. So a bifurcation occurs at a = 1/8.Note that the x coordinate of all the critical points is between 1 and 2.The Jacobian matrix is
J(x, y) =
(2x− 3 1−2ax 1
).
For cases 2 and 3 the trace is clearly positive, and the determinant isdet = 2x(a+ 1)− 3.
11. Writex′ = x(K − x− y), y′ = y(1− x− 2y).
58
a > 1/8 a = 1/8 a < 1/8
. ..
Figure 18: Exercise 9. The three cases a > 1/8 (no critical points), a = 1/8(one critical point), a < 1/8 (two critical points).
The critical points are
(0, 0), (0, 1/2), (K, 0), (2K − 1, 1−K).
Note that the last critical point is viable only when 1/2 ≤ K ≤ 1. Thenullclines are straightforward to sketch (see Figure 19). The Jacobian is
J(x, y) =
(K − 2x− y −x
−y 1− 4y
).
Easily the eigenvalues of J(0, 0) are K and 1, so (0, 0) is an unstable node.The eigenvalues of J(K, 0) are −K and 1, so (K, 0) is a saddle point. Theeigenvalues of J(0, 1/2) are K − 1/2 and −1, so (K, 0) is a saddle point.Finally, trJ(2K − 1, 1 −K) = 2K − 3 < 0 and det trJ(2K − 1, 1 −K) =4(−3K2 + 5K − 2) > 0 on 1/2 < K < 1. Thus, as the direction fieldshows, (2K − 1, 1−K) is a stable node.
The cases K < 1/2 and K = 1/2 are straightforward. The results areshown in Figure 19.
13. The system is
x′ = y, y′ = −y − x− x3 + F cos t.
See page 356 of the text for a MATLAB code.
15. Add the two differential equations and solve the result for x to get x = 1.Substitute into the y equation to get y = 4a. So (1, 4a) is the criticalpoint. The Jacobian evaluated at the critical point is
J(x, y) =
(a− 1 1
4−a −1
4
).
59
K<1/2 K=1/2 1/2< K<1
(1/2,0)(0,0) (0,0)(0,0) (K,0)
(0,1/2)(0,1/2)
(K,0) 1
(0,1/2)
K
K
1 1
Figure 19: Exercise 11. When K < 1/2 and K = 1/2 all orbits approach thepoint (0, 1/2), which is a stable node. For K > 1/2 the orbits approach the thecoexistent state (2K − 1, 1−K).
We have trJ = a − 5/4 and det J = 1/4 > 0. Therefore the criticalpoint is asymptotically stable for a < 5/4 and unstable for a > 5/4.At a = 5/4 the linearized system has a center. Therefore, a bifurcationoccurs at a = 5/4. Compute the discriminant (trJ)2 − 4 det J to observethat (1, 4a) is a stable spiral when a < 5/4 and an unstable spiral whena > 5/4.
A calculation of the phase plane is shown in Figure 20 when a = 3/2.
17. There are two critical points, (0, 0) and (1, 0). The x nullcline is y = 0 (xaxis), and the y nullcline is the parabola y = −1
cx(1 − x). The Jacobianmatrix is
J(x, y) =
(0 1
2x− 1 −c
).
Substituting the two critical points into J instantly gives the fact that(0, 0) is a stable node (both eigenvalues are negative) and (1, 0) is a saddlepoint (negative determinant). Figure 21 shows the direction field. Theseparatrix leaving (0,0) in the region between the nullclines starts to theleft and crosses the parabolic nullcline horizontally. At that instant itmust turn NW and enter the origin. It can be shown by calculation thatit cannot cross the straight line y = λ+x, where (1, λ+) is the eigenvectorcorresponding to the eigenvalue
λ+ =1
2(−c+
√c2 − 4).
Note that the other eigenpair is
λ− =1
2(−c−
√c2 − 4), (1, λ−).
An orbit connecting two critical points is called heteroclinic.
60
x ’ = 1 − (a + 1) x + .25 y x2
y ’ = a x − 0.25 y x2 a = 6/4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
1
2
3
4
5
6
7
x
y
Figure 20: Exercise 15. The phase plane diagram when a = 3/2 and the criticalpoint is an unstable spiral. The figure was produced by pplane8, a programwritten by John Polking of Rice University. The program is available on theinternet.
61
(0,0) (1,0)
ll +-
x
y
Figure 21: Exercise 17. Critical points, nullclines and direction field. The twolines labeled λ+ and λ− are eigen-directions for the two negative eigenvalues atthe origin. The orbit connecting the two critical points (1, 0) to (0, 0) (shown)is trapped to the right of the direction λ+ and must enter the origin.
19. Assume x, y ≥ 0. The x nullclines are x = 0 and y = x(1− x), a concave-down parabola. The y nullclines are y = 0 and x = 1/α. The criticalpoints are (0, 0) and (1, 0) for α ≤ 1. When α > 1 there is another criticalpoint at
P =
(1
α,α− 1
α2
).
Figure 22 shows the cases. The Jacobian is
J(x, y) =
(2x− 3x2 − y −x
y x− 1α
).
J(0, 0) has eigenvalues 0, −1/α, and J(1, 0) has eigenvalues −1, 1− 1/α.
a=1a<1a>1
Figure 22: Exercise 19.
62
