Solutions to Problem Set 2
1.
Part A:
dP
 0.08P  2400
dt
P(0)  20,000
P(t )  ce 0.08t  30000
P(0)  c  30000  20000  c  50000
P(t )  50000e 0.08t  30000
P(18)  181,034.79
Part B:
Note: We will be paying an extra $200 a month, so $2400 a year. After 5.625
years we will have paid back the $13,500.
The model for time = [0,5.625] is
dP
 0.08 P  4800
dt
P (0)  6,500
P(t )  ce 0.08t  60000
P(0)  c  60000  6500  c  66500
P(t )  66500e 0.08t  60000
P(5.625)  44,292.76
At 5.625 years the model reverts back to the model in part A.
dP
 0.08P  2400
dt
P(5.625)  44,292.76
P(t )  ce 0.08t  30000
P(0)  ce 0.08*5.625  30000  44292.76  c  47371.15
P(t )  47371.15e 0.08t  30000
P(18)  169,939.24
Part C:
The difference is that in part A from the beginning you are accumulating interest
on $20,000 (that really adds up). We are cheating our son out of $11,095.55
2. f ( y)  2 y 3 (2  y 2 ) 4 (3  y) 9 is continuous.
f
 (6 y 2 (2  y 2 ) 4 (3  y ) 9 ) 2 y 3 [9(3  y ) 8 (2  y 2 ) 4  8 y (3  y ) 9 (2  y 2 ) 3 ] is
y
continuous. The conditions of the existence and uniqueness theorem are satisfied.
Therefore there exists exactly one solution. There are equilibrium solutions at
y=0 & y=3. By uniqueness you cannot cross equilibrium solutions. If we start in
between these equilibrium solutions we must stay in between them for all time.
This solution satisfies 0  y (t )  3 for all t and the domain of y (t ) is the whole
line.
3.
4.
dy
 y ( y 2   ) . Equilibrium solutions are 0 &    . When   0 , there is
dt
only one equilibrium point (y=0) and when α<0 there are three equilibrium
solutions. α=0 is the bifurcation value. Connect the points like a pitchfork for the
bifurcation diagram.
α<0
α=0
α>0
dS
1000 S
S


S (0)  5000
dt
50000  1000t
50  t
Separate
1
1
 dS 
dt
S
50  t
 ln | S | ln | 50  t | c
ln | S | 1  ln | 50  t | c
S 1  c(50  t )
1
S
c(50  t )
1
S (0) 
 5000  c  250000
50c
1
S (t ) 
250000(50  t )
The tank is full after 50 minutes. S=2500 pounds making the concentration
S (50)
2500
1


100000 100000 40
5.
The bifurcations occur when α = 1 & -1. α creates a shift up or down of f(y). If α is
greater than one, there is a shift more than 1 unit up and we have one equilibrium point.
If α = 1 there are two equilibrium points. If α is between -1 and 1 then there are 3
equilibrium points. If α = -1 then there are two equilibrium points. If α < -1 then there is
one equilibrium point.
6.
dy
 y  t 3  cos 3t
dt
homogeneous solution
dy
 y
dt
y h (t )  ce   dt  ce t
y p  at 3  bt 2  ct  d  e cos 3t  f sin 3t
dy p
dt
dy p
dt
 3at 2  2bt  c  3e sin 3t  3 f cos 3t
 y p  3at 2  2bt  c  3e sin 3t  3e cos 3t  at 3  bt 2  ct  d  e cos 3t  f sin 3t  t 3  cos 3t
 at 3  (3a  b)t 2  (2b  c)t  (c  d )  (3e  f ) sin 3t  (e  3 f ) cos 3t  t 3  cos 3t
a 1
3a  b  0  b  3
2b  c  0  c  6
 3e  f  0
e3f 1
1
3
, f 
10
10
1
3
y (t )  y h  y p  ce t  t 3  3t 2  6t  6  cos 3t  sin 3t
10
10
1
y (0)  c 
 6  c  5.9
10
1
3
y (t )  5.9e t  t 3  3t 2  6t  6  cos 3t  sin 3t
10
10
c  d  0  d  6
e
7.
dy 3
 y  2t 3 e 2t
dt t
The integrating factor:
3
  dt
t
1
t3
Multiply both sides of the equation by the integrating factor:
1 dy 1 3
1
 3 ( y )  3 2t 3 e 2t
3
t dt t t
t
Rewrite the LHS:
d 1
( 3 y )  2e 2 t
dt t
Integrate both sides and solve for y:
1
y  e 2t  C
3
t
y(t )  t 3 e 2t  Ct 3
e
3
 e 3 ln t  e ln t  t 3 
y(1)  e 2  C  0  C  e 2
y(t )  t 3 e 2t  e 2 t 3