Math 2280-001 Quiz 3 SOLUTIONS January 30, 2015

```Math 2280-001
Quiz 3 SOLUTIONS
January 30, 2015
1) Consider the following differential equation for a function x t , which could be modeling a logistic
population with harvesting:
x# t =Kx2 C 2 x C 3 .
1a) Draw the phase diagram. Identify the equilibrium solutions and whether or not they are stable.
(3 points)
We look for roots of the slope function - they will be the equilibrium points corresponding to constant
(equilibrium) solutions. Then we analyze whether solutions x t are increasing or decreasing between
equilibrium points. Factoring...
Kx2 C 2 x C 3 =K x2 K 2 x K 3 =K x K 3 x C 1 .
So the equilibrium points are x = 3, x =K1.
x !K1 0 x# t = K K K = K 0 x t decreasing;
K1 ! x ! 3 0 x# t = K K C = C 0 x t increasing;
x O 3 0 x# t = K C C = K 0 x t decreaing.
So the phase diagram is
))K1//3))
and x hK1 is unstable; x h 3 is asymptotically stable.
1b) Solve the initial value problem for the differential equation above, with x 0 = 1 :
x# t =Kx2 C 2 x C 3 .
x 0 =1.
Hint: To save time with partial fractions on this very brief quiz you may use the identity
1
1
1
1
=K
K
.
xC1 xK3
4 xC1
xK3
(6 points)
1
Note: On the quiz this partial fractions decomposition was wrong: the minus sign in front of the
was
4
missing. You will not be marked down if you used the incorrect partial fractions decomposition correctly.
separate variables:
x# t =K x K 3 x C 1
dx
=Kdt
xK3 xC1
dx
xK3
1
4
K
= Kdt
xC1
1
1
K
xC1
xK3
dx = Kdt
1
1
K
dx = 4 dt = 4 t C C
xC1
xK3
ln x C 1 K ln x K 3 = 4 t C C
xC1
ln
= 4 tCC
xK3
exponentiate...
xC1
= e4 t C C = eCe4 t = C1 e4 t .
xK3
by continuity,
xC1
= Ce4 t .
xK3
x 0 =10
2
= C =K1.
K2
xC1
=Ke4 t
xK3
x C 1 = x K 3 Ke4 t =Ke4 t x C 3e4 t
x 1 C e4 t =K1 C 3e4 t .
K1 C 3e4 t
x t =K
.
1 C e4 t
1c) For your solution x t to (b), verify that t lim
x t does agree with the value implied by your phase
/N
diagram in part (a).
(1 point)
4t
It's easier to take the limit if we divide the top and bottom by e :
eK4 t C 3
x t =K K4 t
.
e C1
0C3
lim x t =
= 3.
t /N
0C1
This agrees with what the phase diagram said - for any initial values x0 OK1, t lim
x t = 3.
/N
```