MATH 246: Exam 3 (Friday, May 2, 2014)
Please answer each numbered question on a different answer sheet. Make sure
your name, section number, and the appropriate problem number appear on each
answer sheet. You only need to sign the pledge on the first answer sheet. Show all
work and be sure your answers are complete, with explanations where necessary. You
are not allowed any outside materials on this exam.
1. (10 pts) Consider the following MATLAB code:
>> A = [ 1 2 3; 1 2 3; 1 2 3];
>> [U, R] = eig(A)
0 0 0
Given that part of the output is R = 0 0 0 find a possible output for U .
0 0 6
From R we know the eigenvalues are 0 and 6, and 0 has algebraic multiplicity
2. The eigenvectors for 0 are given by the 
equation
x1+ 2x

 2 + 3x3 = 0 so
−2
−3
x1 = −2x2 − 3x3 . The vector solution is then  1  x2 +  0  x3 . The vectors
0
1
 
 
−2
−3
 1  and  0  are the eigenvectors for 0. For 6 you would row reduce A − 6I.
0
1




−5 2
3
1 0 −1
So you would have  1 −4 3 . This is row equivalent to 0 1 −1
1
2 −3
0 0 0
 
1

Solving the equation (A − 6I)x = 0 we would get 1 x3 (Note: you could have
1
also gotten this using equations instead of matrices, or even just noticed that
the sum of each row is 6, so (1,1,1) has to be an eigenvector for 6. This means
−2 −3 1
0 1 .
a possible R is 1
0
1 1
2. Consider the system
x0 = 4x − 5y
y 0 = 2x − 2y
(a) (10 pts) Find a real general solution to the system.
2
First off the characteristic polynomial
is z +
2z+ 2 so the rootsare −1 ± i.
4 −5
−1 + i
0
Then we want to find a solution to
−
x = 0. So
2 −2
0
−1 + i
we have 2x1 = (−3+ i)x2 (or if you prefer (5 − i)x1 = −5x2 ). Then a vector
−3 + i
solution is
. So the real general solution is
2
−t cos(t) − 3 sin(t)
−t −3 cos(t) − sin(t)
.
+ c2 e
c1 e
2 sin(t)
2 cos(t)
(b) (10 pts) Classify the equilibrium solution (0,0).
It is a stable spiral.
(c) (5 pts) Sketch a phase portrait of with all pertinent trajectories.
You should have a spiral winding counterclockwise into the origin.
1 2
3. Consider the matrix A =
.
−2 5
(a) (5 pts) Find the Characteristic Polynomial p(z) for A.
z 2 + 6z + 9
(b) (5 pts) Find a natural set of solutions to p(D)[y] = 0.
−3t
−3t
1 0
}. Then we have
X =I
−3 1
A fundamental set of solutions is {e , te
1 0
so X =
so the natural set of solutions is {e−3t + 3te−3t, te−3t }.
3 1
(c) (5 pts) Find eAt .
This is given by
e−3t + 4te−3t
2te−3t
.
N0 (t)I + N1 (t)A =
−2te−3t
e−3t + 8te−3t
1
(d) (10 pts) Find a solution to the IVP x = Ax where x(0) =
.
2
0
You could do this with generalized eigenvectors, but since you
in theory didall
−3t
e + 8te−3t
the work already you can use eAt . So you would have that
.
2e−3t + 14te−3t
More on the back!!!
4. Suppose that each of the following are L[y] for some y. Find y.
(a) (10 pts)
e−s
s2 + 2s + 1
(c) (10 pts)
(b) (10 pts)
(s2
(a) This is e−s /(s + 1)2 so e−s L[te−t ].
1) ((t − 1)e1−t ).
e2s
s(s2 + 1)
2s
+ 2s + 2)
Then the inverse Laplace is u(t −
(b) First we use partial fractions. We have s(s21+1) = As + Bs+C
. So As2 + A +
s2 +1
Bs2 + Cs = 1. Then A = 1, B = −1 and C = 0. So the expression is equivalent
e2s L[1 + cos(t)]. So the inverse Laplace is
u(t + 2) (1 + cos(t + 2)) .
2s
(c) (s2 +2s+2)
= 2(s+1)−2
= L[2e−t cos(t) − 2e−t sin(t)]. So the inverse Laplace is
(s+1)2 +1
2e−t cos(t) − 2e−t sin(t).
5. (10 pts) Consider the initial value problem y 00 − y = f (t), y(0) = 1, y 0 (0) = 2
where
sin(t) for 0 ≤ t < π
f (t) =
.
0
for t ≥ π
Find an expression for L[y]. You do NOT need to find y!!
First we rewrite f as sin(t) − sin(t)u(t − π). Then taking the Laplace on both
−πs
. So
sides we have s2 Y − s − 2 − Y = 1−e
s2 +1
Y (s) =
(s + 2)(s2 + 1) + 1 − eπs
.
s4 − 1
Check your work! Have a great weekend!!
A Short List of Laplace Transforms
n!
for s > a
(s − a)n+1
s−a
L[eat cos(bt)](s) =
for s > a
(s − a)2 + b2
L[tn eat ](s) =
L[eat sin(bt)](s) =
b
for s > a
(s − a)2 + b2
L[tn f (t)](s) = (−1)n F (n) (s) where F (s) = L[f (t)](s)
L[eat f (t)](s) = F (s − a) where F (s) = L[f (t)](s)
L[u(t−c)f (t−c)](s) = e−cs F (s) where F (s) = L[f (t)](s) and u is the unit step function.