School

University of California, Berkeley **We aren't endorsed by this school

Course

MATH 204

Subject

Mathematics

Date

Nov 20, 2023

Type

Other

Pages

3

Uploaded by HighnessSpider3639 on coursehero.com

Math 22B: Homework 3: Part A
Due: before August 23 at 11:59 pm
Submit on Gradescope
(Make sure to select pages!)
1. Solve the following Bernoulli differential equations:
(a)
y
′
= 5
y
+
e
−
2
x
y
−
2
(b)
dy
dx
−
y
x
=
y
9
(c)
dy
dx
+
2
y
x
=
x
2
y
2
sin(
x
)
(d)
dy
dx
−
2
y
x
=
x
2
y
2
sin(
x
)
(e)
x
dy
dx
−
(1 +
x
)
y
=
xy
2
(f)
dy
dx
=
y
(
xy
3
−
1)
2. The following differential equations are homogeneous. Express them in
the form
dy
dx
=
G
(
y
x
).
Then, make the substitution
v
=
y
x
and solve
them as linear DE.
(a)
dy
dx
=
x
+2
y
x
−
y
(b)
dy
dx
=
x
2
+
y
2
2
x
2
(c)
dy
dx
=
y
(
x
−
y
)
x
2
(d)
x
2
dy
+
y
(
x
+
y
)
dx
= 0
3. Find and classify the equilibrium solutions of the following DE:
(a)
y
′
=
y
2
−
y
−
6
(b)
y
′
= (
y
2
−
4)(
y
+ 1)
2
4. For each of the following, sketch the slope field and phase line, then
identify the equilibrium solutions and classify them asymptotically.
(a)
dy
dt
=
−
3(
y
−
1)
2
(b)
dy
dt
= 3
y
−
7
√
y
(c)
dy
dt
=
y
2
(
y
2
−
1)
(d)
dy
dt
=
y
3
−
3
y
2
+ 2
y
5. Does the following IVP necessarily have to have a unique solution with
the given initial conditions?
dy
dx
=
x
p
y
−
3
(a)
y
(4) = 3
(b)
y
(
−
2) = 28
6. Consider the IVP
y
′
=
10
3
xy
2
/
5
,
y
(0) = 1
.
It ha a unique solution on some open interval containing
x
0
= 0. Find
the solution and determine the largest open interval on which it is
unique.
1

7. Suppose that the populations
y
of a certain species of fish in a given
area of the ocean is described by the logistic equation
dy
dt
=
r
(1
−
y
K
)
y.
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. The
next few problems explore some of the questions involved in formulating
a rational strategy for managing the fishery.
(a) 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/time, that measures the total
effort made to harvest the given species of fish.
To include this
effect, the logistic equation is modified and replaced by
dy
dt
=
r
(1
−
y
K
)
y
−
Ey.
This equation is known as the Schaefer model after the biologist
M. B. Schaefer, who applied it to fish populations.
i. Show that if
E < r
, then there are two equilibrium points,
y
1
= 0 and
y
2
=
K
(1
−
E/r
)
>
0.
ii. Show that
y
1
is unstable and
y
2
is asymptotically stable.
iii. 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
y
2
. Find
Y
as a function
of the effort
E
; the graph of this function is known as the
yield-effort curve.
iv. Determine
E
so as to maximize
Y
and thereby find the maxi-
mum sustainable yield
Y
m
.
(b) In this problem we assume that fish are caught at a constant rate
h
independent of the size of the fish population. Then
y
satisfies
dy
dx
=
r
(1
−
y/K
)
y
−
h.
The assumption of a constant catch rate
h
may be reasonable when
y
is large but becomes less so when
y
is small.
i. If
h < rK/
4, show we have two equilibrium points
y
1
and
y
2
with
y
1
< y
2
; determine these points.
ii. Show that
y
1
is unstable and
y
2
is asymptotically stable.
iii. From a plot of
f
(
y
) versus
y
, show that if the initial population
y
0
> y
1
, then
y
→
y
2
and
t
→ ∞
, but that if
y
0
< y
1
, then
y
decreases as
t
increases. increases. Note that
y
= 0 is not
an equilibrium point, so if
y
0
< y
1
, then extinction will be
reached in a finite time.
2

iv. If
h > rK/
4, show that
y
decreases to zero as
t
increases,
regardless of the value of
y
0
.
v. If
h
=
rK/
4, show that there is a single equilibrium point
y
=
K/
2 and that this point is semistable (In cases where
solutions on one side of an equilibrium solution move towards
the equilibrium solution and on the other side of the equi-
librium solution move away from it we call the equilibrium
solution semi-stable). Thus the maximum sustainable yield is
h
m
=
rK/
4, corresponding to the equilibrium value
y
=
K/
2.
Observe that
h
m
has the same value as
Y
m
in Part (a). The
fishery is considered to be overexploited if
y
is reduced to a
level below
K/
2.
Note: this last one is from the textbook:
section 2.5 problems 19 and 20.
3