Ord Diff Eqs
HW#01 Solutions
Page 1 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
1) Draw a direction field for y  3  2 y , determine the behavior as t   , and state how this behavior may
depend on the initial value of y at t = 0.
Look for an equilibrium solution by taking:
y  0
0  3 2y
3
2
From the direction field, it appears that the slopes are negative for y > 1.5, so any initial value y(0) > 1.5
converges to y = 1.5. Also, slopes appear to be positive for y < 1.5, so any initial value y(0) < 1.5 also
converges to y = 1.5. So, as t   , the behavior of all solutions is to converge to y = 1.5.
y
Ord Diff Eqs
HW#01 Solutions
Page 2 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
5) Draw a direction field for y   1  2 y , determine the behavior as t   , and state how this behavior may
depend on the initial value of y at t = 0.
Look for an equilibrium solution by taking:
y  0
0  1 2 y
1
2
From the direction field, it appears that the slopes are positive for y > -0.5, so any initial value y(0) > -0.5
diverges away from y = 1.5. Also, slopes appear to be negative for y < -0.5, so any initial value y(0) < -0.5 also
diverges from y = 1.5. So, as t   , the behavior of all solutions is to diverge away from the equilibrium
solution y = -0.5.
y
Ord Diff Eqs
HW#01 Solutions
Page 3 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
7) Write a differential equation of the form
dy
 ay  b whose solution has the behavior that all solutions
dt
approach y = 3 as t   .
 Choose a < 0 for convergence, and for the correct value of the equilibrium solution, we need:
dy
0
dt
0  ay  b
b
y 3
a
dy
 y 3
 one choice is
dt
Ord Diff Eqs
HW#01 Solutions
Page 4 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
9) Write a differential equation of the form
dy
 ay  b whose solution has the behavior that all other solutions
dt
diverge from y = 2 as t   .
 Choose a > 0 for divergence, and for the correct value of the unstable equilibrium solution, we need:
dy
0
dt
0  ay  b
b
y 2
a
dy
 y2
 one choice is
dt
Ord Diff Eqs
HW#01 Solutions
Page 5 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
11) Draw a direction field for y  y  4  y  , determine the behavior as t   , and state how this behavior may
depend on the initial value of y at t = 0.
Look for equilibrium solutions by taking:
y  0
0  y 4  y
y  0, y  4
From the direction field, it appears that for initial conditions y(0) > 0, all solutions converge to the equilibrium
solution of y = 4. For initial conditions y(0) < 0 all solutions appear to diverge to -.
Ord Diff Eqs
HW#01 Solutions
Page 6 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
13) Draw a direction field for y  y 2 , determine the behavior as t   , and state how this behavior may depend
on the initial value of y at t = 0.
Look for equilibrium solutions by taking:
y  0
0  y2
y  0 (double root)
From the direction field, it appears that for initial conditions y(0) < 0, all solutions converge to the equilibrium
solution of y = 0. For initial conditions y(0) > 0 all solutions appear to diverge to +.
15) Stable equilibrium at y = 2  this is equation (j) y  2  y
Ord Diff Eqs
HW#01 Solutions
Page 7 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
16) Unstable equilibrium at y = 2  this is equation (c) y  y  2
17) Stable equilibrium at y = -2  this is equation (g) y  2  y
Ord Diff Eqs
HW#01 Solutions
Page 8 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
18) Unstable equilibrium at y = -2  this is equation (b) y  2  y
19) Unstable equilibrium at y = 0 and stable equilibrium at y = 3  this is equation (h) y  y  3  y 
Ord Diff Eqs
HW#01 Solutions
Page 9 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
20) Unstable equilibrium at y = 3 and stable equilibrium at y = 0  this is equation (e) y  y  y  3
21) A pond initially contains 1,000,000 gallons of an undesirable chemical; water with 0.01 g of this chemical
flows in at a rate of 300 gal/hr. Water flows out at the same rate, so the volume of water remains constant (and
the chemical mixes instantaneously throughout the pond)
(a) Differential equation for amount of chemical in pond at any time:
Let M(t) be the amount of chemical (measured in grams) in the pond as a function of time t (measured in hours)
 0.01g   300 gal 
g
 The amount of chemical that flows into the pond is 

3
hr
 gal   hr 
 The amount of the chemical that flows out will be the concentration of the chemical times the volume that
 M  t    300 gal 
3
flows out:  6

  4 M t 
 10 gal   hr  10 hr
 The rate of change of the chemical will be what flows in minus what flows out:
M t 
dM
g
 3  0.0003
dt
hr
hr
Ord Diff Eqs
HW#01 Solutions
Page 10 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
(b) The amount of chemical in the pond after a very long time will be:
dM
0
dt
M t 
0  0.0003
hr
g
3
hr  104 g
M t    
0.0003
hr
From the direction field, it appears that this will be the amount after a very long time no matter how much is in
the pond to begin with:
22) Write a differential equation for the volume of a raindrop if it evaporates at a rate proportional to its surface
area.
 Let V(t) be the volume of the drop as a function of time, and let A(t) be the area as a function of time.
2
1/3
  3V 1/3 
4
1/3
2/3
 3V 
2
A

4

r
V   r3  r  
A

4

;

 

    4   3V 
3
 4 
  4  
dV
 kA for k > 0 
 The differential equation is
dt
dV
1/3
2/3
 k  4   3V 
dt
dV
 aV 2/3
dt
where the constant a combines the constant k and the numeric constants.
Ord Diff Eqs
HW#01 Solutions
Page 11 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
25) For rapidly falling bodies, it is more correct to model the air drag as being proportional to the swuare of the
velocity.
(a) Write a differential equation:
dv
 Instead of m  mg   v . We would now have:
dt
dv
m  mg   v 2
dt
dv

 g  v2
dt
m
(b) Limiting velocity:
dv
0
dt
0g

m
vterm.2
mg
vterm. 

(c) If m = 10 kg, find drag coefficient so that limiting velocity is 49 m/sec
0g

m
mg

vterm.2
v2
10kg   9.8m / s 2 

2
 49m / s 

2 kg
kg
 0.0408
49 m
m
Ord Diff Eqs
HW#01 Solutions
Page 12 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
(d) Draw a direction field and compare with linear drag:
dv

 g  v2
dt
m
2 kg
m 49 m 2
 9.8 2 
v
s
10kg
dv
m
1
 9.8 2 
v2
dt
s 245m
Ord Diff Eqs
HW#01 Solutions
Page 13 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
27) Draw a direction field for y  te2t  2 y , determine the behavior as t   , and state how this behavior may
depend on the initial value of y at t = 0.
All solutions appear to approach the equilibrium solution of y = 0.
Ord Diff Eqs
HW#01 Solutions
Page 14 of 14
Section 1.1: #1, 5, 7, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 22, 25, 27, and 31.
31) Draw a direction field for y  2t  1  y 2 , determine the behavior as t   , and state how this behavior may
depend on the initial value of y at t = 0.
Look at:
y  0
0  2t  1  y 2
y   2t  1
It appears that for y   2t  1 , all solutions converge toward something like y  2t  1 , while for
y   2t  1 , all solutions diverge to -.