ODE Homework 1
2.1. Linear Equations; Method of Integrating Factors
1. Find the solution of the given initial value problem
cos t
2
y 0 + y = 2 , y(π) = 0, t > 0
t
t
[§2.1 #16]
Sol. The integrating factor µ(t) is
Z
2
dt = e2 ln |t| = t2
µ(t) = exp
t
Multiple the equation by µ(t), we get
t2 y 0 + 2ty = (t2 y)0 = cos t
Integrating both side of the equation, we have that
sin t
+ Ct−2
t2
for some constant C. Since y(π) = 0, so
t2 y = sin t + C ⇒ y(t) =
C
sin π
+ 2 ⇒C=0
2
π
π
Hence the specific solution of the initial value problem is
0=
y(t) =
sin t
t2
2. Find the solution of the given initial value problem
ty 0 + (t + 1)y = t,
y(ln 2) = 1,
t>0
[§2.1 #20]
Sol. Rewrite the equation as the form
t+1
y0 +
y=1
t
The integrating factor µ(t) is
Z
t+1
µ(t) = exp
dt = et+ln |t| = tet
t
since t > 0. Multiple the equation by µ(t), we get
tet y 0 + (t + 1)et y = (tet y)0 = tet
Integrating both side of the equation, we have that
Z
t − 1 Ce−t
t
te y = tet dt = (t − 1)et + C ⇒ y(t) =
+
t
t
for some constant C. Since y(ln 2) = 1, so
ln 2 − 1 Ce− ln 2
+
⇒C=2
1=
ln 2
ln 2
Hence the specific solution of the initial value problem is
y(t) =
t + 2e−t − 1
t
3. Consider the initial value problem
1
y 0 + y = 3 + 2 cos 2t, y(0) = 0.
4
(a) Find the solution of this initial value problem and describe
its behavior for large t.
(b) Determine the value of t for which the solution first intersects the line y = 12.
[§2.1 #29]
Sol.
R 1 t
(a) The integrating factor µ(t) = exp
dt = e 4 . And the
4
equation can be written as
t
t
t
t
1 t
e 4 y 0 + e 4 y = (e 4 y)0 = 3e 4 + 2e 4 cos 2t
4
Integrating both side, we can conclude that the general
solution for the equation is
t
8
64
y(t) = Ce− 4 +
cos 2t +
sin 2t + 12
65
65
Together with the initial condition y(0) = 0, we get the
solution for the initial value problem is
788 − t
8
64
y(t) = −
e 4+
cos 2t +
sin 2t + 12
65
65
65
788 − t
8
=−
e 4 + √ sin(2t + θ) + 12
65
65
where θ = sin−1
64
.
65
As t → ∞, we have that
8
y(t) → √ sin(2t + θ) + 12
65
which means the solution will oscillate about an average
12,with an amplitude of √865 .
(b) By letting y(t) = 12, then we have the equality
t
8 cos 2t + 64 sin 2t − 788e− 4 = 0
t
Let g(t) = 8 cos 2t + 64 sin 2t − 788e− 4 , then g(10) ≈
−2.99 < 0, g(10.3) ≈ 1.53 > 0. By using Newton’s method
with initial point x0 = 10, we have that the first value t for
which y(t) intersects y = 12 is t ≈ 10.066.
4. Consider the initial value problem
3
y 0 − y = 3t + 2et , y(0) = y0
2
Find the value of y0 that separates solutions that grow positively
as t → ∞ from those that grow negatively. How does the
solution that corresponds to this critical value of y0 behave as
t → ∞?
[§2.1 #31]
R 3 3t
− 2 dt = e− 2 . And
Sol. The integrating factor µ(t) = exp
the equation can be written as
−3t
3t
3t
t
3 −3t
e− 2 y 0 − e 2 y = (e 2 y)0 = 3te− 2 + 2e− 2
2
Integrating both side, we have that the general solution for the
equation is
3t
4
y(t) = Ce 2 − 4et − 2t −
3
The initial condition y(0) = y0 implies that the specific solution
for the initial value problem is
16 3t
4
y(t) = y0 +
e 2 − 4et − 2t −
3
3
3t
e 2 will dominate the solution. Its
As t → ∞, the term y0 + 16
3
sign will determine the divergence properties. Hence the critical
value of the initial condition is y0 = − 16
. The corresponding
3
4
t
solution y(t) = −4e − 2t − 3 , will also decrease without bound.
5. Show that all solutions of 2y 0 + ty = 2 approach a limit as
t → ∞, and find the limiting value.
[§2.1 #32]
Proof. Rewrite the equation as y 0 + 2t y = 1. The integrating
R t t2
4 .
factor µ(t) = exp
dt
=
e
And the equation can be
2
written as
t2
t2 0
t2
t t2
e 4 y0 + e 4 y = e 4 y = e 4
2
Integrating both side, we have that the general solution for the
equation is
Z
t
e
y(t) =
s2 −t2
4
ds
Thus, by L’Hospital Rule,
R t s2
Z t 2 2
e 4 ds
s −t
lim y(t) = lim
e 4 ds = lim
t2
t→∞
t→∞
t→∞
e4
t2
= lim
e4
t→∞ t
e
2
t2
4
2
=0
t→∞ t
= lim
Hence we are done.
6. Variation of Parameters. Consider the following method of
solving the general linear equation of first order:
y 0 + p(t)y = g(t).
(a) If g(t) = 0 for all t, show that the solution is
h Z
i
y = A exp − p(t)dt ,
(i)
(ii)
where A is a constant.
(b) If g(t) is not everywhere zero, assume that the solution of
Eq. (i) is of the form
h Z
i
y = A(t) exp − p(t)dt ,
(iii)
where A is now a function of t. By substituting for y in
the given differential equation, show that A(t) must satisfy
the condition
hZ
i
0
A (t) = g(t) exp
p(t)dt .
(iv)
(c) Find A(t) from Eq. (iv). Then substitute for A(t) in Eq.
(iii) and determine y. Verify that the solution obtained
in this manner agrees with that of Eq. (33) in the text.
This technique is known as the method of variation of
parameters.
[§2.1 #38]
Proof.
0
(a) If g(t) = 0 for all t, then the equation becomes
=
R t y + p(t)y
0. The integrating factor is µ(t) = exp
p(s)ds . And
the equation can be written as
hZ t
i
hZ t
hZ t
i 0
0
exp
p(s)ds y +p(t) exp
p(s)ds y = exp
p(s)ds y = 0
Integrating both side, we have that the general solution for
the equation is
h Z t
i
y(t) = A exp −
p(s)ds
for some constant A.
(b) For the case g(t) 6≡ 0, assume that
the
of Eq.
R t solution
(i) is of the form y(t) = A(t) exp − p(s)ds . Since
y 0 + p(t)y = g(t), so
h Z t
i
0
A (t) exp −
p(s)ds = g(t)
Hence
0
A (t) = g(t) exp
hZ
t
i
p(s)ds
Rt
(c) Since A0 (t) = g(t) exp
p(s)ds , we have that
Z t
Z t
hZ s
i
Rs
A(t) =
g(s) exp
p(τ )dτ ds =
g(s)e p(τ )dτ ds
Hence
h
Z
t
i Z t
Rs
Rt
g(s)e p(τ )dτ ds · e− p(s)ds
p(s)ds =
y(t) = A(t) exp −
Z t
Rs
=
g(s)e t p(τ )dτ ds
Rt
By letting µ(t) = exp
p(s)ds , then the expression of
y(t) is exactly the same with Eq. (33) in the text book.
7. Use the method of variation of parameters to solve the given
differential equation.
ty 0 + 2y = cos t,
t>0
[§2.1 #41]
Sol. Rewrite the equation as Rthe form
y 0 + 2t y = cost t . The
2
integrating factor µ(t) = exp
dt = t2 . Hence the general
t
solution of the homogeneous equation y 0 + 2t y = 0 is
h Z 2 i A
y0 (t) = A exp −
dt = 2
t
t
for some constant A. Assume that the solution for the original
. Substituting y(t) into the
equation is of the form y(t) = A(t)
t2
equation we get
cos t
A0 (t) =
µ(t) = t cos t
t
R
and thus A(t) = t cos tdt = t sin t+cos t+C for some constant
C. Therefore, the solution y(t) of the equation is
y(t) =
t sin t + cos t + C
A(t)
=
2
t
t2
2.2. Separable Equations
8. Solve the given differential equation.
dy
x − e−x
=
dx
y + ey
[§2.2 #7]
Sol. Rewrite the equation of the form (y + ey )dy = (x − e−x )dx.
Then
Z
Z
y2
x2
y
(y + e )dy = (x − e−x )dx ⇒ ey +
= e−x +
+C
2
2
for some constant C, which defines the solution y(x) of the
equation implicitly.
9. Solve the initial value problem
3x2
, y(1) = 0
3y 2 − 4
and determine the interval in which the solution is valid.
[§2.2 #22]
Sol. Rewrite the equation of the form (3y 2 − 4)dy = 3x2 dx.
Then
Z
Z
2
(3y − 4)dy = 3 x2 dx ⇒ y 3 − 4y = x3 + C
y0 =
for some constant C, which defines the general solution y(x)
of the equation implicitly. Since y(1) = 0, so we have that
1 + C = 0 ⇒ C = −1. Hence the solution y(x) for the initial
value problem is defined implicitly by the equation
y 3 − 4y = x3 − 1
2
Note that the differential equation y 0 = 3y3x2 −4 implies that the
slope of tangent lines tends to infinity as y → ± √23 . Substitut√ . By
ing y = ± √23 into y 3 − 4y = x3 − 1 we get x3 − 1 = ∓ 316
3
solving above two equations, we have that x ≈ −1.276 and 1.598
respectively. Therefore, the solution y(x) is valid for about the
interval (−1.276, 1.598).
10. Solve the initial value problem
y0 =
2 cos 2x
,
3 + 2y
y(0) = −1
and determine where the solution attains its maximum value.
[§2.2 #25]
Sol. Rewrite the equation of the form (3 + 2y)dy = 2 cos 2xdx.
Then
Z
Z
(3 + 2y)dy = 2 cos 2xdx ⇒ y 2 + 3y = sin 2x + C
for some constant C, which defines the general solution y(x) of
the equation implicitly. Since y(0) = −1, so have that C = −2
so that y 2 +3y = sin 2x−2. Furthermore, the quadratic formula
together with the initial condition y(0) = −1 implies that
r
3
1
y(x) = − + sin 2x +
2
4
Note that y 0 (x) = √ cos 2x
sin 2x+ 41
, we have that the solution is valid
for sin 2x > − 14 , together with the initial condition y(0) = −1,
the solution is valid on interval I = sin−1 (− 14 ), π2 −sin−1 (− 14 ) .
Thus x = π4 is the only critical point in the interval I. Observe
that
π 4
= −√ < 0
y 00
4
5
√
Then y(x) has maximum at x = π4 , which is y π4 = 5−3
2
11. Consider the initial value problem
y0 =
ty(5 − y)
,
1+t
y(0) = y0 > 0.
(a) Determine how the solution behaves as t → ∞.
(b) If y0 = 2, find the time T at which the solution first reaches
the value 4.99.
(c) Find the range of initial values for which the solution lies
in the interval 4.99 < y < 5.01 by the time t = 2.
[§2.2 #28]
Sol.
dy
tdt
(a) Rewrite the equation of the form y(5−y)
= 1+t
. Then
Z
Z
dy
tdt
=
⇒ ln |y| − ln |5 − y| = 5t − 5 ln |1 + t| + C0
y(5 − y)
1+t
y ⇒ ln = 5t − 5 ln |1 + t| + C0
5−y
y Ce5t
⇒
=
5−y
(1 + t)5
|y0 |
where C = |5−y
according to the initial condition y(0) =
0|
y0 > 0. As t → ∞, we have that
y Ce5t
55 Ce5t
= lim
=∞
→ lim
t→∞ (1 + t)5
t→∞
5−y
5!
y 5
=
Hence 5−y
− 1 → ∞ which implies that y(t) → 5
5−y
as t → ∞.
2
(b) If y(0) = 2, then C = 5−2
= 23 , and hence the solution for
the initial value problem is defined implicitly by
y
2e5t
=
5−y
3(1 + t)5
2e5t
10te5t
Since dtd 3(1+t)
= 3(1+t)
5
6 ≥ 0 for all t ≥ 0, we have that
y
> 0 for all t ≥ 0. According to the differential equation
5−y
itself, we know that y 0 is always positive and hence y(t) is
increasing for all t ≥ 0. To obtain T , we simply solve the
equation
4.99
2e5T
=
5 − 4.99
3(1 + T )5
to get T ≈ 2.6063.
(c) Note that y(t) = 5 is an equilibrium solution. According
to (b), we know that if y0 < 5, then y(t) < 5, ∀ t ≥ 0 and
similarly, if y0 > 5, then y(t) > 5, ∀ t ≥ 0. Hence that the
general solution for the equation is defined implicitly by
y
y0 e5t
=
,
5−y
(5 − y0 )(1 + t)5
y0 6= 5
y
y
< −499 or 5−y
>
For t = 2, if 4.99 < y < 5.01, then 5−y
501. Thus we have that
y0
499 · 35
501 · 35
y0
< − 10 or
>
5 − y0
e
5 − y0
e10
This implies that the initial value y0 lies about approximately in the interval (4.23397, 6.10986).
12. Consider the equation
y − 4x
dy
=
.
dx
x−y
(a) Show that Eq. (i) can be rewritten as
dy
(y/x) − 4
=
;
dx
1 − (y/x)
(b)
(c)
(d)
(e)
(f)
(i)
(ii)
thus Eq. (i) is homogeneous.
Introduce a new dependent variable v so that v = y/x, or
dy
dv
y = xv(x). Express dx
in terms of x, v and dx
.
dy
Replace y and dx in Eq. (ii) by the expressions from part
dv
(b) that involve v and dx
. Show that the resulting differential equation is
dv
v−4
v+x
=
,
dx
1−v
or
dv
v2 − 4
x
=
,
(iii)
dx
1−v
Observe that Eq. (iii) is separable.
Solve Eq. (iii), obtaining v implicitly in terms of x.
Find the solution of Eq. (i) by replacing v by y/x in the
solution in part (d).
Draw a direction field and some integral curves of Eq.(i).
Recall that the right side of Eq. (i) actually depends only
on the ratio y/x. This means that integral curves have the
same slope at all points on any given straight line through
the origin, although the slope changes from one line to another. Therefore the direction field and the integral curves
are symmetric with respect to the origin. Is this symmetry
property evident from your plot?
[§2.2 #30]
Sol.
(a) By dividing x on both the numerator and denominator of
the right hand side of Eq. (i), we get Eq. (ii).
dy
dv
(b) Since y = vx, so dx
= v + x dx
.
(y/x)−4
dy
v−4
(c) Since dx = 1−(y/x) = 1−v , by (b), we get
dv
v−4
dv
v2 − 4
=
⇒x
=
dx
1−v
dx
1−v
(d) Note that Eq. (iii) is an separable equation, we can rewrite
dx
the equation as the form v1−v
2 −4 dv = x . Then
Z
Z
1−v
dx
dv =
⇒ ln |2 − v| + 3 ln |2 + v| = −4 ln |x| + C0
2
v −4
x
1
⇒ ln |(2 − v)(2 + v)3 | = ln 4 + C0
x
⇒ x4 |v − 2||v + 2|3 = C
v+x
where C = eC0 for some constant C0 . Thus the general
solution v(x) for Eq. (iii) is defined implicitly by the equation
x4 |v − 2||v + 2|3 = C
(e) Note that y = vx, so the general solution y(x) of Eq. (i) is
defined implicitly by
(y − 2x)(y + 2x)3 = C ⇒ y 4 + 4xy 3 − 16x3 y − 16x4 = C
(f) The plot of the direction field and integral curves of Eq.
(i) is as follows
13. Given differential equation
(x2 + 3xy + y 2 )dx − x2 dy = 0
(a) Show that the given equation is homogeneous.
(b) Solve the differential equation.
(c) Draw a direction field and some integral curves. Are they
symmetric with respect to the origin?
[§2.2 #36]
Sol.
(a) Rewrite the equation as the form
y y 2
dy
x2 + 3xy + y 2
=
=
1
+
3
+
dx
x2
x
x
which implies that the equation is homogeneous.
(b) Let v = xy , then the above equation becomes
dv
dv
= 1 + 3v + v 2 ⇒ x
= (v + 1)2
dx
dx
dv
dx
Thus, (v+1)
2 = x , and hence
Z
Z
dv
dx
−1
=
⇒
= ln |x| + C
2
(v + 1)
x
v+1
1
−1
⇒v=
C − ln |x|
v+x
Replacing v = xy , we got the general solution y(x) for the
x
original equation that y = C−ln
− x.
|x|
(c) The plot of the direction field and integral curves for the
equation is as follows Clearly, the integral curves are sym-
metric with respect to the origin.
2.5. Autonomous Equations and Population Dynamics
14. Consider the equation of the form dy
= f (y), y0 ≥ 0, where
dt
f (y) = y(y − 1)(y − 2). Sketch the graph of f (y) versus y.
Determine the critical (equilibrium) points, and classify it as
asymptotically stable or unstable. Draw the phase line, and
sketch several graphs of solutions in the ty-plane.
[§2.5 #3]
Sol. The graph of f (y) is as follows. Since the critical points
occurs at dy
= 0, it is easy to see that y = 0, 1, 2 are critical
dt
points. That is, y = 0, 1, 2 are equilibrium solutions for the
equation. Note that f (y) is positive on (0, 1) ∪ (2, ∞), while it
is negative on (−∞, 0) ∪ (1, 2). We see that the solution y(t) is
increasing for 0 < y < 1 and y > 0, while decreasing for y < 0
and 1 < y < 2. Hence the critical points y = 0 and y = 2 are
unstable, while the critical point y = 1 is asymptotically stable.
The graph of solutions and the phase line is as follows
15. Consider the equation of the form dy
= f (y), −∞ < y0 < ∞,
dt
2 2
where f (y) = y (y − 1). Sketch the graph of f (y) versus y.
Determine the critical (equilibrium) points, and classify it as
asymptotically stable or unstable. Draw the phase line, and
sketch several graphs of solutions in the ty-plane.
[§2.5 #9]
Sol. The graph of f (y) is as follows
Since the critical points occurs at dy
= 0, it is easy to see that
dt
y = 0, ±1 are critical points. That is, y = 0, ±1 are equilibrium solutions for the equation. Note that f (y) is positive on
(−∞, −1) ∪ (1, ∞), while it is negative on (−1, 0) ∪ (0, 1). We
see that the solution y(t) is increasing for y < −1 and y > 1,
while decreasing for −1 < y < 0 and 0 < y < 1. Hence the
critical points y = 0 is semistable, while the critical point y = 1
is unstable, and the critical point y = −1 is asymptotically stable. The graph of solutions and the phase line is as follows:
16. Suppose that a certain population obeys the logistic equation
dy
= ry 1 − Ky .
dt
(a) If y0 = K4 , find the time τ at which the initial population
has doubled. Find the value of τ corresponding to r =
0.025 per year.
)
(b) If yK0 = α, find the time T at which y(T
= β, where 0 <
K
α, β < 1. Observe that T → ∞ as α → 0 or as β → 1.
Find the value of T for r = 0.025 per year, α = 0.1, and
β = 0.9.
[§2.5 #15]
Sol.
(a) Observe that the logistic equation is separable. By solving
the equation, we get the general solution is
y(t) =
y0 K
y0 + (K − y0 )e−rt
where y0 = y(0) is the initial condition. We can write t as
a function of y,
−1 y0 (K − y) t=
ln r
y(K − y0 )
K−y . Setting y =
With y0 = K4 , we have that t = −1
ln
r
3y
2y0 = K2 , we get
τ=
−1 K − K2 ln 3
ln 3K =
r
r
2
If r = 0.025 per year. τ ≈ 43.9445 year.
)
(b) By letting α = yK0 and β = y(T
= Ky , then
K
−1 α(1 − β) T =
ln r
β(1 − α)
If r = 0.0025 per year and α = 0.1, β = 0.9, then T ≈
175.778 year.
17. Another equation that has been used to model population growth
is the Gompertz equation
K dy
= ry ln
,
dt
y
where r and K are positive constants.
(a) Sketch the graph of f (y) versus y, find the critical points,
and determine whether each is asymptotically stable or
unstable.
(b) For 0 ≤ y ≤ K, determine where the graph of y versus t is
concave up and where it is concave down.
(c) For each 0 < y ≤ K, show that dy
as given by the Gomdt
dy
pertz equation is never less than dt as given by the logistic
equation.
[§2.5 #16]
Sol.
(a) Let f (y) = ry ln Ky , then f 0 (y) = r ln Ky − 1 . It is
clear that f (0) = f (K) = 0 and f 0 (y) = 0 if and only if
y = Ke < K. The graph of f (y) versus y is as follows
The critical points are y = 0 and y = K. Since f (y) is
positive on (0, K) and negative on (K, ∞), the solution y(t)
is increasing for 0 < y < K and is decreasing for y > K.
Thus y = 0 is unstable and y = K is asymptotically stable.
(b) Note that for 0 ≤ y ≤ K, dy
= f (y) > 0. By chain rule,
dt
d2 y
d dy d
dy
=
= f (y) = f 0 (y)
2
dt
dt dt
dt
dt
Observe that if 0 ≤ y ≤ Ke , f (y) is increasing while if
K
≤ y ≤ K, f (y) is decreasing. Thus f 0 (y) > 0 for 0 ≤
e
y ≤ Ke and f 0 (y) < 0 for Ke ≤ y ≤ K. This implies that
the solution y(t) is concave up for 0 ≤ y ≤ Ke while it is
concave down for Ke ≤ y ≤ K.
(c) Let g(y) = ry 1 −
h(K) = 0, and
h0 (y) =
y
K
and set h(y) =
f (y)−g(y)
.
ry
Then
yf 0 (y) − yg 0 (y) − f (y) + g(y)
1
1
− ≤0
=
2
ry
K y
since 0 < y ≤ K. This shows that
f (y) ≥ g(y) for dy0 <
K
y ≤ K. Therefore, dy
=
ry
ln
is never less than dt =
dt
y
y
ry 1 − K for 0 < y ≤ K.
18. Harvesting a Renewable Resource. Suppose that the papulation y of a certain species of fish (for example, tuna or halibut)
in a given area of the ocean is described by the logistic equation
dy
y
=r 1−
y.
dt
K
Although it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish
population may be reduced below a useful level and possibly
even driven to extinction. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends
on the population y: the more fish there are, the easier it is to
catch them. Thus we assume that the rate at which fish are
caught is given by Ey, where E is a positive constant, with
units of 1/times, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic
equation is replaced by
dy
y
=r 1−
y − Ey.
dt
K
This equation is known as the Schaefer model after the biologist M. B. Schaefer, who applied it to fish populations.
(a) Show that if E < r, then there are two equilibrium points
y1 = 0 and y2 = K 1 − Er > 0.
(b) Show that y = y1 is unstable and y = y2 is asymptotically
stable.
(c) A sustainable yield Y of the fishery is a rate at which fish
can be caught indefinitely. It is the product of the effort
E and the asymptotically stable population y2 . Find Y
as a function of the effort E; the graph of this function is
known as the yield-effort curve.
(d) Determine E so as to maximize Y and thereby find the
maximum sustainable yield Ym .
[§2.5 #20]
Sol.
(a) Rewrite the equation of the form
dy
ry =y r−E−
dt
K
dy
Then dt = 0 if and only if y = 0 or
ry
E
r−E−
=0⇒y =K 1−
K
r
Hence if E < r, then there
are two equilibrium points
E
y1 = 0 and y2 = K 1 − r > 0.
(b) Let f (y) = y r − E − ry
, then f (y) > 0 for 0 < y < y2
K
and f (y) < 0 for y > y2 . This implies that the solution
y(t) is increasing for 0 < y < y2 and is decreasing for
y > y2 . Therefore y = y1 = 0 is unstable and y = y2 is
asymptotically stable.
(c) It is easy to see that
E
Y = y2 E = KE 1 −
r
(d) Note that
KE 2
K
r 2 rK
Y = KE −
=−
E−
+
r
r
2
4
r
Hence when E = 2 , we have maximum sustainable yield
Ym =
rK
4
19. Bifurcation Points. For an equation of the form
dy
= f (a, y),
(i)
dt
where a is a real parameter, the critical points (equilibrium
solutions) usually depend on the value of a. As a steadily increases or decreases, it often happens that at a certain value of
a, called a bifurcation point, critical points come together,
or separate, and equilibrium solutions may either be lost or
gained. Bifurcation points are a great interest in many applications, because near them the nature of the solution of the underlying differential equation is undergoing an abrupt change.
For example, in fluid mechanics a smooth (laminar) flow may
break up an become turbulent. Or an axially loaded column
may suddenly buckle and exhibit a large lateral displacement.
Or, as the amount of one of the chemicals in a certain mixture
is increased, spiral wave patterns of varying color may suddenly
September 11, 2008 11:18 emerge
boyce-9e-bvp
number 113
Page number
93 The
cyan black
in an Sheet
originally
quiescent
fluid.
following exercise
consider the equation
dy
= a − y2.
(ii)
dt
2.5 Autonomous Equations and Population Dynamics
93
(a) Find all of the critical points for Eq. (ii). Observe that
there are no critical points if a < 0, one critical point if
fluid. Problems
through
describe three
typesifofabifurcations
a = 0, 25and
two27critical
points
> 0. that can occur in simple
equations of the form (i).
(b) Draw the phase line in each case and determine whether
25. Consider the equation
each critical point is asymptotically
stable, semistable, (ii)
or
dy/dt = a − y2 .
unstable.
(a) Find all of the critical points for Eq. (ii). Observe that there are no critical points if
(c)
each
case
several
solutions
Eq. (ii) in the tya <In
0, one
critical
pointsketch
if a = 0, and
two critical
points if aof
> 0.
(b) plane.
Draw the phase line in each case and determine whether each critical point is asymptotically stable, semistable, or unstable.
(d)
If we plot the location of the critical points as a function of
(c) In each case sketch several solutions of Eq. (ii) in the ty-plane.
a
in the ay-plane, we obtain Figure 2.5.10. This is called
(d) If we plot the location of the critical points as a function of a in the ay-plane, we obtain
the2.5.10.
bifurcation
diagram
Eq.for(ii).
The
at0
Figure
This is called the
bifurcation for
diagram
Eq. (ii).
The bifurcation
bifurcation at a =
is called
This name is more
natural in theThis
contextname
of second
a =a0saddle–node
is calledbifurcation.
a saddle-node
bifurcation.
is
order systems, which are discussed in Chapter 9.
more natural in the context of second order systems.
y
2
1
–2
–1
Asymptotically stable
1
–1
2
3
4 a
Unstable
–2
FIGURE 2.5.10
Bifurcation diagram for y = a − y2 .
[§2.5 #25]
Sol.
dy/dt = ay − y3 = y(a − y2 ).
(iii)
dy
(a) Let dx = 0, we have y 2 = a. Then there are no critical
(a) Again consider the cases a < 0, a = 0, and a > 0. In each case find the critical points,
points
if line,
a <and0,determine
there iswhether
one each
critical
= 0 if a =
0,
draw
the phase
criticalpoint
point
stable,
√ isyasymptotically
semistable,
or
unstable.
and there are two critical points y = ± a if a > 0.
(b) For
In each
case0,sketch
several
solutions
of Eq.point.
(iii) in the
ty-plane.
(b)
a<
there
is no
critical
For
a = 0, the critical
(c) Draw the bifurcation diagram for Eq. (iii), that is, plot the location of the critical points
point
y
=
0
is
semistable.
For
a
>
0
the√ critical
point
versus a. For
bifurcation;
your
√ Eq. (iii) the bifurcation point at a = 0 is called a pitchfork
diagram
why this name is appropriate.
y = mayasuggest
is asymptotically
stable and y = − a is unstable.
26. Consider the equation
27. Consider the equation
dy/dt = ay − y2 = y(a − y).
(iv)
(a) Again consider the cases a < 0, a = 0, and a > 0. In each case find the critical points,
draw the phase line, and determine whether each critical point is asymptotically stable,
semistable, or unstable.
(b) In each case sketch several solutions of Eq. (iv) in the ty-plane.
The phase lines of each cases is as follows
a<0
a=0
a>0
(c) The graph of solutions for each case is as follows
a<0
a=0
a>0