Solutions

MATB44 Assignment 1 Question 3: solution
(3) Determine a model for the population P (t) if both the birth and the
death rate are proportional to the population present at time t.
Solution:
The birth rate is
βP (t).
The death rate is
δP (t).
The birth rate is the number of births per unit time per unit population.
So
dP
= (β − δ)P 2 (t)
dt
This equation is separable:
1 dP
== β − δ
P 2 dt
so
−
1
= (β − δ)t − C
P (t)
or
P (t) =
1
.
C + (δ − β)t
Here C is a constant.
1
Solution to Assignment 1 questions 7 and 8
Question 7:
y ′ = −1 − y/x
This equation is homogeneous.
Set v = y/x
So
dy/dx = v + xdv/dx = −1 − v
dv
= −1 − 2v
dx
dv
dx
=
−1 − 2v
x
x
Integrating, we obtain
1
− ln |1 + 2v| = ln |x| + C1
2
Taking the exponential,
|1 + 2v|−1/2 = C2 x
Squaring and taking the reciprocal,
1 + 2v = C3 (1/x2 )
or
x + 2y
C3
= 2
x
x
C3
x + 2y =
x
or
C4 x
− .
x
2
Here C1 , . . . , C4 are constants. (C4 = C3 /2, C3 = (C2 )−2 and C2 = eC1 .)
y=
1
Question 8:
(x2 − 3y 2 )dx + 2xydy = 0, y(2) = 1
This equation is homogeneous.
Substitute v = y/x
dy/dx = −(x2 − 3y 2)/2xy = −(1 − 3v 2 )/2v
dv
v+x
= −1/v + 3/2v
dx
dv
x
= −1/v + 1/2v
dx
dx
2vdv
=
2
−2 + v
x
Sub w = v 2
dx
dw
=
−2 + w
x
ln |w − 2| = ln |x| + C
ln |v 2 − 2| = ln |x| + C
ln |y 2/x2 − 2| = ln |x| + C
This is the general solution (in implicit form).
Taking the exponential,
y 2 /x2 − 2 = |x|eC
y 2 /x2 = 2 + |x|eC
y 2 = (2 + |x|eC )x2
Introducing the initial condition y(2) = 1,
ln |1/4 − 2| = ln 2 + C
so
C = ln(7/4) − ln 2.
eC = (7/4) × (1/2) = 7/8.
y 2 = (2 + 7/8|x|)x2
2