Math 53
Midterm II
May 13, 2008
Name:
(please print clearly)
Signature:
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Instructions:
• No calculators, books, notes, or electronic devices may be used during the exam.
• You have 90 minutes.
• There are 7 problems, some with multiple parts. Complete all problems, and write all solutions
in the space provided on the exam. Please be sure to indicate your intended answer clearly,
and cross out any work that you do not want considered.
• Additional scratch paper is provided at the end of the exam.
• Before beginning the exam, please indicate your discussion section (circle one):
Andrés Angel:
10:00am (53-10)
1:15pm (53-09)
Tania Rojas Esponda:
10:00am (53-06)
11:00am (53-05)
Eric Schoenfeld:
1:15pm (53-03)
2:15pm (53-12)
Kamil Szczegot:
11:00am (53-08)
2:15pm (53-11)
Problem 1 (10 points)
Problem 2 (10 points)
Problem 3 (10 points)
Problem 4 (10 points)
Problem 5 (10 points)
Problem 6 (10 points)
Problem 7 (10 points)
Total (70 points)
1. Find the general solution to the system
5 1
x =
x.
−1 3
0
5 1
. The characteristic polynomial of A is
First we find the eigenvalues of A =
−1 3
5−λ
1
= λ2 − 8λ + 16 = (λ − 4)2 .
Det
−1 3 − λ
Hence the matrix has only one eigenvalue, λ = 4.
Second we find the eigenvectors of A with eigenvalue 4. These are described by the matrix
equation
1
1
v1
0
=
.
−1 −1 v2
0
This systems of equations is equivalent to v 1 + v2 = 0.
1
We take v =
.
−1
1
(Any multiple of
is a fine choice for v.)
−1
Since we have only a 1-dimensional space of eigenvectors, we need to look for a generalized
eigenvector of A. This is a solution to the equation (A−4I)w = v. Writing out this equation,
we get
1
1 w1 , w2 = 1, −1 .
−1 −1
This system of equations is equivalent to w 1 + w2 = 1.
1
We take w =
.
0
(Any other vector w with w1 + w2 = 1 is a fine choice for w. Also, if you chose a different
eigenvector v in the last step, you will end up with a different w in this step.)
The general solution for a 2x2 system with a repeated eigenvalue is
c1 eλt v + c2 eλt [tv + w].
In our case, the general solution is
1
1
1
4t
4t
c1 e
+ c2 e [t
+
].
−1
−1
0
1
2. Find the general solution to the following second order equation
y 00 + 4y 0 + 4y = sin 2t.
First we find a particular solution using the method of undetermined coefficients. Since the
right-hand side is a sine, we consider combinations of sines and cosines.
We try y = A sin 2t + B cos 2t.
So y 0 = 2A cos 2t − 2B sin 2t.
And y 00 = −4A sin 2t − 4B cos 2t.
Hence y 00 + 4y 0 + 4y = −4A sin 2t − 4B cos 2t + 8A cos 2t − 8B sin 2t + 4A sin 2t + 4B cos 2t.
Grouping terms and simplifying, we see that
y 00 + 4y 0 + 4y = −8B sin 2t + 8A cos 2t.
Hence y solves our equation if B = −1/8 and A = 0.
We have found a particular solution Y = −(1/8) cos 2t.
Second we find the general solution of the homogeneous equation y 00 + 4y 0 + 4y = 0. The
characteristic polynomial is λ2 + 4λ + 4 = (λ + 2)2 . The polynomial has a double root at
λ = −2. Therefore, the general solution to the homogeneous equation is c 1 e−2t + c2 te−2t .
Combining these two steps, we see that the general solution of our equation is
y = (−1/8) cos 2t + c1 e−2t + c2 te−2t .
2
3. For each of the following 2 × 2 linear systems, find all equilibrium solutions, write down the
general (real) solution, and then sketch a phase portrait. Be sure to include arrows on your
diagram that indicate the direction of the trajectories. (In each case, we’ve provided you with
eigenvectors of the matrix appearing in the system, so you do not need to compute these).
4
1
0
2 −4
0
and
Eigenvectors:
x+
(a) x =
1
2
−3
2 −7
Equilibrium solution(s):
−2
x(t) =
−1
General solution:
1
4
−2
−6t
t
x(t) = c1 e
+ c2 e
+
2
1
−1
First we find
solution by solving
theequilibrium
2 −4 e1
0
+
= 0.
2 −7 e2
−3
Either writing out the equations or using row-reduction,
we get e 1 = −2 and e2 = −1.
2
−4
x.
Second we solve the homogeneous system x 0 =
2 −7
We are given the eigenvectors of the matrix. To find the eigenvalues, multiply the matrix
by each
eigenvector
and see what happens.
2−4 1
−6
=
.
2−7 2
−12
1
is an eigenvector with eigenvalue −6.
Therefore,
2
2−4 4
4
=
.
2−7 1
1
4
Therefore,
is an eigenvector with eigenvalue 1.
1
Assembling all of the information we’ve found so far, we can write down the general
solution. −2
1
4
−6t
t
.
x = c1 e
+ c2 e
+
−1
2
1
Notes about drawing the solution: The solution is centered at the equilibrium point
(−2, −1)T . Because one eigenvalue is positive and one is negative it is a saddle type
drawing.
3
(problem 3 continued)
3 1
±i
0
(b) x =
x
Eigenvectors:
−1 3
1
Equilibrium solution(s):
x(t) = 0
General (real) solution:
sin t
cos t
3t
3t
x(t) = c1 e
+ c2 e
cos t
− sin t
The equilibrium solutions are the solutions to the equation
3 1 e1
= 0.
−1 3 e2
Solving this system, we see that the only equilibrium solution is 0.
i
Next we solve the system. We are given a complex eigenvector
. To find the corre1
sponding eigenvalue, we multiply the matrix by the eigenvector.
3 1 i
1 + 3i
i
=
= (3 − i)
.
−1 3 1
3−i
1
i
Therefore,
is an eigenvector with eigenvalue (3 − i).
1
i
(3−i)t
We now have a complex solution to our differential equations, z = e
. To get
1
real solutions, we take the real and imaginary parts of z.
i
i
i
sin t
cos t
e(3−i)t
= e3t (cos(−t)+i sin(−t))
= e3t (cos t−i sin t)
= e3t
+ie3t
.
1
1
1
cos t
− sin t
Therefore the general real solution of our equation is
sin t
cos t
3t
3t
c1 e
+ c2 e
.
cos t
− sin t
Notes about drawing the solution: Our system has eigenvalues 3 ± i. The eigenvalues
are complex and the real part is positive. Therefore, the solutions spiral outwards.
From the eigenvalues alone, we cannot tell whether the solutions spiral clockwise or
counterclockwise. To figure that out, we write down one of the solutions and examine it
more carefully.
sin t
3t
.
x(t) = e
cos t
4
0
, a point on the positive y-axis. The solution makes one full
At time t = 0, x(0) =
1
revolution around the origin
in time
3π/2
2π, crossing the positive y axis again at t = 2π. At
e
time π/2, the solution is
on the positive x-axis. At time π, the solution is on
0
the negative y-axis, and at time 3π/2, the solution is on the negative x-axis. From this
information, we deduce that the solution is moving clockwise.
Because the eigenvalues are 3 ± i, each solution moves outward quite a lot by the time
it makes one full revolution around the origin. The solution we examined above starts
at (0, 1)T on the positive y-axis at time 0. It hits the positive y-axis again at time 2π,
at the location (0, e6π ). Note that e6π is quite large, roughly 108 . Because of this, the
curves that you see in the picture only curve slightly and don’t visibly spiral around.
5
4. Find the general solution to the system


1
3 1
x0 =  0
2 0 x.
−1 −2 1


−1
[To save you some time, you may assume −2 is an eigenvector of the above matrix with
5
eigenvalue 2.]
First we find the characteristic polynomial of the matrix.


1−λ
3
1
Det  0
2−λ
0  = −λ3 + 4λ2 − 6λ + 4.
−1
−2 1 − λ
Therefore the eigenvalues are the roots of λ 3 − 4λ2 + 6λ − 4. We already know that 2 is an
eigenvalue of this matrix. Therefore, we can factor a (λ − 2) out of the polynomial, and we
get the following.
λ3 − 4λ2 + 6λ − 4 = (λ − 2)(λ2 − 2λ + 2).
Therefore, the other two eigenvalues are the roots of λ 2 − 2λ + 2, which are λ = 1 ± i.
We already know the eigenvector corresponding to the eigenvalue 2.
Next we find the eigenvector corresponding to the eigenvalue 1 + i. This vector is a solution
of the equation

   
0
−i
3
1
v1
 0 1 − i 0  v2  = 0.
0
v3
−1 −2 −i
We solve this system by row-reducing. There are many ways to do this. I’m going to carry
out one way. As we go, I’ll describe in words what we’re doing and discuss strategy a little
bit. The first thing I’ll do is move the bottom row to the top and multiply it by −1 to get a
1 in the upper left-hand corner.

 

−i
3
1
1
2
i
 0 1 − i 0  ∼  0 1 − i 0 .
−1 −2 −i
−i
3
1
Next, I’m going to use the 1 in the upper left-hand corner to kill all the other terms in the
left column. One of those terms is already zero, so I only have to add i times the top row to
the bottom row.


1
2
i
∼ 0 1 − i 0 .
0 3 + 2i 0
6
Next I want to deal with the second column. I can take the second row and divide it by (1−i)
to get (0, 1, 0) in that row.


1
2
i
∼ 0
1
0 .
0 3 + 2i 0
Then
other

1
∼ 0
0
by adding appropriate multiple of the second row to the other rows, I can kill all the
terms in the second column.

0 i
1 0 .
0 0
I notice that the bottom row ended up being zero. This confirms that 1 + i is an eigenvalue
and is a good check that I haven’t messed up the computation.
Finally, I get the two equations v1 + iv3 = 0 and v2 = 0.
 
−i

I choose v = 0 .
1
 
1
(Any multiple of this vector would be a fine choice. For example, you may have chosen 0.)
i
At this point, we may either look for real solutions or complex solutions. If we look for real
solutions we would proceed as follows.
 
 




−i
−i
sin t
− cos t
e(1+i)t  0  = et (cos t + i sin t)  0  = et  0  + iet  0  .
1
1
cos t
sin t
Hence we get two real solutions corresponding to the complex eigenvalue 1+i, and the general
real solution of the ODE is




 
− cos t
sin t
−1
c1 e2t −2 + c2 et  0  + c3 et  0  .
sin t
cos t
5
If we look for complex solutions, we must use BOTH complex eigenvalues. Since 1 − i is the
complex conjugate of 1 + i, we can take the eigenvector for 1 − i to be the complex conjugate
of the eigenvector we had for 1 + i.
Hence the general complex solution is the following.
 
 
 
i
−i
−1
(1−i)t
(1+i)t
2t





0 .
0 + c3 e
−2 + c2 e
c1 e
1
1
5
Bear in mind that in the general real solution, the constants c 1 , c2 , and c3 are real, while in
the general complex solution they are complex.
7
5. Suppose that

2e2t
x1 (t) =  −e2t 
−2e2t


−e2t
x2 (t) =  8e2t 
−5e2t


e2t
x3 (t) = −3e2t 
e2t

are solutions to a homogeneous, constant-coefficient, 3 × 3 linear system x 0 = Ax. Can the
general solution to the system be determined using this information?
Since x1 (t), x2 (t), and x3 (t) are each solutions of our equation x 0 = Ax, it follows that any
linear combination of them is a solution. In other words, c 1 x1 + c2 x2 + c3 x3 is a solution to
our equation for any constants c1 , c2 , and c3 .
Next we check whether we have found all solutions of our equation. In particular, we check
whether the solutions above include all possible initial values at time t = 0. The initial values
of the three solutions are x1 (0) = (2, −1, −2)T , x2 (0) = (−1, 8, −5)T , and x3 (0) = (1, −3, 1)T .
We want to check whether these three vectors span the whole 3-dimensional space. To do
that, we check whether the vectors are linearly independent by taking the determinant of the
matrix with columns x1 (0), x2 (0), and x3 (0).


2 −1 1
DET −1 8 −3 =
−2 −5 1
2[8 − (−3)(−5)] − (−1)[−1 − (−3)(−2)] + 1[(−1)(−5) − 8(−2)]
= 2[8 − 15] + [−1 − 6] + [5 + 16] = −14 − 7 + 21 = 0.
Since the determinant is zero, the vectors are linearly dependent.
The set of solutions c1 x1 + c2 x2 + c3 x3 does not include all initial values, and so they are
not all of the solutions to the ODE. Hence we cannot determine the general solution to the
differential equation.
We didn’t ask you to find the linear dependence, but with a little more work you can find
that
x1 − x2 − 3x3 = 0.
Therefore, x3 is a linear combination of x1 and x2 : x3 = (1/3)x1 − (1/3)x2 .
8
6. Consider the initial value problem
y 00 + 5y 0 + 4y = 0
y(0) = 1
y 0 (0) = v.
For which values of v does y cross 0 at some time t ≥ 0?
We begin by finding the general solution to the equation y 00 + 5y 0 + 4y = 0. The characteristic
polynomial is λ2 + 5λ + 4 = (λ + 1)(λ + 4). The roots are −4 and −1. Therefore, the general
solution is
c1 e−t + c2 e−4t .
The constants c1 and c2 depend on the initial conditions. The initial conditions give us the
following two equations.
1 = y(0) = c1 + c2 .
v = y 0 (0) = −c1 − 4c2 .
We solve the equations for c1 and c2 in terms of v.
From the first equation, c2 = 1 − c1 . We plug this into the second equation.
v = −c1 − 4(1 − c1 ) = 3c1 − 4.
c1 = (v + 4)/3.
c2 = (−v − 1)/3.
Hence the solution to our differential equation is
y(t) = (1/3)(v + 4)e−t − (1/3)(v + 1)e−4t .
We want to know whether y(t) = 0 for some time t ≥ 0. I will explain two ways to check this.
Either approach is fine.
Algebraic approach:
To figure out whether y crosses zero at a positive time, we locate all the times t when
y(t) = (1/3)(v + 4)e−t − (1/3)(v + 1)e−4t = 0.
Manipulating this equation gives
(v + 4)e−t = (v + 1)e−4t .
If v + 4 6= 0 we can divide by v + 4 to get the following equation.
e3t = (v + 1)/(v + 4).
(∗)
If (v + 1)/(v + 4) ≤ 0 then (∗) does not hold for any t, which means that the solution never
crosses zero.
If 0 < (v + 1)/(v + 4), then the solution vanishes at exactly one time t 0 = (1/3) log[(v +
1)/(v + 4)].
9
The time t0 is negative if (v +1)/(v +4) < 1, and the time t 0 is at least 0 if (v +1)/(v +4) ≥ 1.
Finally, if v + 4 = 0, then our solution is y(t) = e −4t which does not cross zero.
We have determined that the solution crosses zero at a time t ≥ 0 if and only if (v+1)/(v+4) ≥
1.
We can simplify the algebra by saying that the solution crosses zero at a time t ≥ 0 if and
only if v < −4.
———Geometric approach:
We want to know whether y(t) = 0 at some time t ≥ 0. Recall that we found the following
formula for y(t).
y(t) = c1 e−t + c2 e−4t = (1/3)(v + 4)e−t − (1/3)(v + 1)e−4t .
We know that c1 + c2 = 1. The behavior of y(t) depends on the signs of c 1 and c2 .
If c1 ≥ 0 and c2 ≥ 0, then y(t) > 0 for all t.
If c1 < 0 then we must have c2 > 0. In this case, the first term c1 e−t is negative and the
second term c2 e−4t is positive. When t is large, e−t has a much larger magnitude than e−4t .
Therefore, the first term dominates, and y(t) < 0 when t is very large. Since y(0) = 1, the
solution must cross zero at some positive time.
If c2 < 0 then we must have c1 > 0. In this case, the first term is positive and the second
term is negative. Since e−t ≥ e−4t , the first term again dominates and the sum is positive.
More precisely, if t ≥ 0, c1 e−t + c2 e−4t ≥ c1 e−4t + c2 e−4t = (c1 + c2 )e−4t = e−4t > 0.
So we see that the solution crosses zero at a positive time if and only if c 1 is negative. Since
c1 = (1/3)(v + 4), the solution crosses zero at a positive time if and only if v < −4.
10
7. Consider the nonlinear second order equation
y 00 = −y − y 2 .
(a) Prove that E = 21 (y 0 )2 + 12 y 2 + 31 y 3 is a conserved quantity for this equation.
To check that E is conserved, we compute E 0 .
E 0 = y 0 y 00 + yy 0 + y 2 y 0 .
Next we use the differential equation, substituting y 00 = −y − y 2 .
= y 0 (−y − y 2 ) + yy 0 + y 2 y 0 = 0.
Since E 0 (t) = 0, E(t) is constant, and so E is a conserved quantity.
(b) Assume a solution y satisfies the initial conditions y(0) = 0, y 0 (0) = 10. If at some time
t1 we know that y(t1 ) = 1, what are possible values of y 0 (t1 )?
At time t = 0, E(0) = (1/2)102 + (1/2)02 + (1/3)03 = 50.
Since E is a conserved quantity, E(t 1 ) = 50 also.
Hence 50 = (1/2)(y 0 (t1 ))2 + (1/2)12 + (1/3)13 .
Rearranging this equation, we see that
y 0 (t1 )2 = 295/3.
Therefore, y 0 (t1 ) = ±(295/3)1/2 .
Based on the information given, the sign of y 0 (t1 ) could be either positive or negative.
(c) Assuming still that y satisfies the initial conditions y(0) = 0, y 0 (0) = 10, is it possible
that y(t) = 100 at some t?
Since y(0) = 0, and y 0 (0) = 10, we have E(0) = 50 as above. Since E is conserved, we
have E(t) = 50 at every time t. If y(t) = 100, then we would have
50 = (1/2)(y 0 (t))2 + (1/2)1002 + (1/3)1003 .
This is impossible, because (1/2)(y 0 (t))2 ≥ 0 and (1/2)1002 + (1/3)1003 is greater than
50.
11