Math 53 - Spring 2009 - Second Midterm Exam
Name:
Student ID:
Signature:
Section Leader
(circle one)
David
Ayala
Kaiyuan
Zhang
Lan
Huang
Kamil
Szczegot
Section time
(circle one)
1:15
2:15
10:00
11:00
11:00
1:15
10:00
2:15
Instructions: Print your name, student ID number, mark your discussion section, and write your
signature to indicate that you accept the honor code. During the test, you may use books and
notes, and you may not use computers, phones, or any other electronic device. Read each question
carefully, and show all your work. Justify all your answers. You have 120 minutes to answer all the
questions. At most two bonus points may be awarded for quality of the presentation.
Question
Score
Maximum
1
10
2
15
3
20
4
15
5
20
6
20
Total
100
Problem 1.
a) (5 pts.) Solve the system of equations
x(0) = 2 and y(0) = 0.
dx
dy
= x + 5y,
= 5y + x with the initial condition
dt
dt
Solution:
The system’s matrix is


 1 5 

A=


5 1
whose eigenvalues satisfy (λ − 1)2 − 25 = 0, hence λ1 = 6 and λ2 = −4. Corresponding
eigenvectors are
 


 1 
 1 



V1 = 
  and V2 = 

1
−1
so that the solutions to the ODE are


6t
4t
 c1 e + c2 e 

 .


c1 e6t − c2 e4t
The initial condition yields c1 + c2 = 2 and c1 = c2 so that c1 = c2 = 1 and x(t) = e6t + e4t ,
y(t) = e6t − e4t .
b) (5 pts.) Draw the phase portrait.
Solution:
There are two straight lines containing solutions, directed by eigenvectors V1 , V2 (the solution
is outgoing on the first, ingoing on the second). Because eigenvalues have opposite signs, the
other trajectories are hyperbola shaped.
1
Problem 2.

a) (10 pts.) Give the formula for the general solution of

1 −1/2 
dQ 
 Q.
=


dt
2
1
Remarks:
• your final answer must be expressed in terms of real functions
• it is unnecessary to justify the formula.
Solution:


 1 −1/2 
, we first obtain the eigenvalues by solving det(A − λI) = (1 −
Letting A = 


2
1
λ)2 + 1 = 0, obtaining λ = 1 + i and λ̄ = 1 − i.
The eigenspace for λ = 1 + i is the nullspace of


 −i −1/2 
,
A − (1 + i)I = 


2
−i


1

which is spanned by the eigenvector V = 

−2i

. A complex solution is given by Q = eλt V,

and we now use the fact that its real and imaginary parts give two linearly independent real
solutions.







eλt V = e(1+i)t 


1
−2i



 = et (cos t + i sin t) 



1

 cos t + i sin t 
 = et 




−2i
2(sin t − i cos t)

 cos t 
 sin t 
 + i et 

= et 




2 sin t
−2 cos t
Hence the general solution of the system is




 sin t 
 cos t 
 + c2 et 
.
c1 et 




−2 cos t
2 sin t
2
b) (5 pts.) Sketch a phase portrait of this system, and state the stability of the critical point
at the origin.
Solution:
The eigenvalues are 1±i, complex with positive real part, hence the phase portrait has a single
critical point at the origin which is unstable, and the phase portrait
  is an outwards spiral. To
 1 

determine its direction, we can for example substitute Q = 
  and find the direction of
0
 
the trajectory through this point. We obtain there
 1 
dQ

=
 , i.e. the trajectory points
dt
2
up and right across the positive x-axis. Hence it is an anticlockwise outwards spiral.
3
dy
Problem 3. Consider the equation y + (3x + 4y) dx
= 0.
a) (5 pts.)
exact.
Find a real number α such that after multiplying this equation by y α it becomes
Solution:
Multiplying by y α gives
y α+1 + y α (3x + 4y)
dy
= 0,
dx
dy
which is of the form M (x, y) + N (x, y) dx
= 0, where M (x, y) = y α+1 and N (x, y) = y α (3x +
∂N
4y). This will be exact when ∂M
∂y = ∂x , i.e.
(α + 1)y α = 3y α .
This equality holds for all y if and only if α = 2; so the equation is exact precisely for α = 2.
b) (10 pts.)
Find the general solution of this equation. (You can leave your answer as an
implicit relation between x and y.)
Solution:
Setting α = 2, our equation becomes
y 3 + y 2 (3x + 4y)
dy
= 0.
dx
3
We will find a function F (x, y) such that ∂F
∂x = M = y and
∂F
3
∂x = y we obtain
F (x, y) = xy 3 + h(y)
∂F
∂y
for some function h of y. Substituting this expression for F into
= N = y 2 (3x + 4y). From
∂F
∂y
= y 2 (3x + 4y) gives
3xy 2 + h0 (y) = 3xy 2 + 4y 3 ,
and hence h0 (y) = 4y 3 . We may take h(y) = y 4 and then we have F (x, y) = xy 3 + y 4 as
our desired function. The solution curves of our differential equation are the level curves of
F (x, y) = y 3 (x + y), i.e. y 3 (x + y) = c for constants c.
1. c) (5 pts.) Solve the equation explicitly when y(1) = −1.
Solution:
The above constant c is c = 0, which implies y = 0 or y = −x. Only the latter corresponds
to our initial condition.
4
Problem 4. Consider the autonomous equation
dy
dt
= ey (y 2 − 3y + 2).
a) (5 pts.) Find the equilibrium points of this system and classify them as stable or unstable.
Solution:
Write f (y) := ey (y 2 − 3y + 2) = ey (y − 1)(y − 2). The equilibria occur when
or y = 2 since ey never vanishes.
dy
dt
= 0 so y = 1
To determine the stability of these equilibria, one can sketch a direction graph, or recall that,
0
at an equilibrium point of dy
dt = f (y), the equilibrium is stable if f (y) < 0 and unstable if
f 0 (y) > 0. Here we have
√ !
√ !
1− 5
1+ 5
0
y 2
y
y−
.
f (y) = e (y − y − 1) = e y −
2
2
We obtain f 0 (1) = −e < 0 and f 0 (2) = e2 > 0, so y = 1 is stable and y = 2 is unstable.
b) (5 pts.) For which values of y is the solution concave up (a.k.a. convex) ? and concave
down ? (Recall that a function y is concave up if y 00 ≥ 0 and concave down if y 00 ≤ 0.)
Solution:
Using chain rule
d2 y
d
dy
= f (y) = f 0 (y)
= f 0 (y)f (y) = e2y (y − 1)(y − 2)(y 2 − y − 1)
2
dt
dt
dt
The zeros of f 0 (y)f (y) are known and it is easy to check that they are
2.
√
1− 5
2
<1<
√
1+ 5
2
<
c) (5 pts.) Draw a qualitatively correct picture of y(t) versus t in some typical cases. (This
question can be answered without solving question b, however information from question b,
when known, can and should be used to refine the picture).
5
Problem 5. Let A be a 2 × 2 matrix, V, W two linearly independent vectors, and λ ∈ R be such
that AV = λV and AW = λW + V.
a) (5 pts.) Write down two solutions Q1 , Q2 of the ODE dQ/dt = AQ, in terms of λ, t, V, W,
such that neither can be written as a linear multiple of the other—in other words, they are
linearly independent. (Linear independence has to be justified properly.)
Solution:
Two possible (and standard) linearly independent solutions are
Q1 (t) = eλt V,
Q2 (t) = eλt (tV + W) .
To see they are linearly independent, suppose some linear combination of them is zero, i.e.
c1 eλt V + c2 eλt (tV + W) = 0.
Since eλt 6= 0 and collecting the V and W terms, we have
(c1 + c2 t) V + c2 W = 0.
Since V and W are linearly independent, this gives c1 + c2 t = 0 and c2 = 0. Substituting
c2 = 0 into the first equation gives c1 = 0 also. Hence Q1 (t), Q2 (t) are linearly independent
for all t. One can also use the Wronskian from the next question.


↑
 ↑


b) (10 pts.) Let Q1 , Q2 be the two solutions from (a). Let W(t) = det 
 Q1 (t) Q2 (t)


↓
↓



 be



the Wronskian of the two solutions. Compute W(t) and express your answer in terms of λ, t
and W(0).
Solution:






 c 
 a 
 a 
 and W =  . Then we have Q1 (t) = eλt   and Q2 (t) =
Let V = 
 
 
 
d
b
b


 ta + c 
. Thus
eλt 


tb + d
a ta + c
eλt a eλt (ta + c)
= e2λt (ad − bc)
= e2λt det W(t) = λt
b tb + d
e b eλt (tb + d)
We wish to express this in terms of W(0), which is W(0) = ad − bc according to the previous
computation. So that W(t) = e2λt W(0).
We could equivalently write, using the bilinear properties of the determinant:
W(t) = det(eλt V, eλt (tV + W)) = e2λt det(V, tV + W) = e2λt det(V, W) = e2λt W(0).
6
c) (5 pts.) Conclude that the Wronskian W never vanishes.
Solution:
Because exponential is never zero, the Wronskian W(t) can vanish iff W(0) = det(V, W) =
ad − bc = 0. But that would imply that V and W are linearly dependent, which contradicts
the hypotheses.
d) [Bonus question] (2 pts.) Show that whenever two non-zero vectors satisfy AV = λV and
AW = λW + V, they are linearly independent.
Solution:
Suppose by contradiction that V, W are dependent, so that there exists a constant α such
that W = αV. Then
AW = A(αV) = αAV = αλV = λαV + V.
So that V = 0, which contradicts the hypothesis V 6= 0.
7
Problem 6.
a) (5 pts.) Determine all critical points of the system



 dx/dt = x − y + 1
.


 dy/dt = x − y + xy − 1
Solution:
We solve dx/dt = dy/dt = 0. First dx/dt = 0 forces y = x + 1, which we plug in the other
equation
0=
dy
= x − (x + 1) + x(x + 1) − 1 = x2 + x − 2 = (x − 1)(x + 2)
dt
whose roots are 1, −2. So that the critical points are A = (1, 2), B = (−2, −1).
b) (15 pts.) For each critical point, find the corresponding linear system, its eigenvalues and
characterize its type (e.g. source, sink, etc.).
Solution:
• For point (1, 2) set x = 1 + u, y = 2 + v and



 du/dt = u − v


 dv/dt = 3u + uv = u + η(u, v)


1 −1
.
with linear matrix A = 


3 0
√
The eigenvalues are imaginary: (1 ± i 11)/2; because of the positive real part, we have
a spiral source.
• For point (−2, −1) set x = −2 + u, y = −1 + v and





 du/dt = u − v
1 −1

with linear matrix A = 




 dv/dt = −3v + uv = u + η(u, v)
0 −3
with (obvious) eigenvalues 1 and −3. So that the equilibrium is a saddle (hence unstable)
8