Download Exam 3 Sol

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Fundamental theorem of algebra wikipedia, lookup

System of polynomial equations wikipedia, lookup

Elementary algebra wikipedia, lookup

System of linear equations wikipedia, lookup

Singular-value decomposition wikipedia, lookup

Eigenvalues and eigenvectors wikipedia, lookup

Linear algebra wikipedia, lookup

History of algebra wikipedia, lookup

Equation wikipedia, lookup

Quartic function wikipedia, lookup

Basis (linear algebra) wikipedia, lookup

Perron–Frobenius theorem wikipedia, lookup

Jordan normal form wikipedia, lookup

Transcript
Solution: APPM 2360
Exam 3
Fall 2011
Problem 1: (20 points) Consider the linear differential operator
2
2
L(y) = y 00 − y 0 + 2 y
t
t
2
(a) Verify that y(t) = t and y(t) = t are both solutions to the homogeneous equation L(y) = 0.
(b) Construct a particular solution to the nonhomogeneous equation L(y) = f , where f (t) = t3 .
(c) Write down the general solution to the equation L(y) = f .
Solution:
(a)
2
2
2
2
L(t) = (t)00 − (t)0 + 2 (t) = 0 − (1) + 2 (t) = 0
t
t
t
t
and
2
2
2
2
L(t2 ) = (t2 )00 − (t2 )0 + 2 (t2 ) = 2 − (2t) + 2 (t2 ) = 0
t
t
t
t
.
(b) Variation of parameters: Two homogeneous solutions given by y1 = t, y2 = t2 .
y1 y2 t t2 = t2 .
W [y1, y2] = 0
=
y1 y20 1 2t
−t2 t3
−y2 f
=
= −t3
W [y1 , y2 ]
t2
y1 f
t · t3
v20 =
= 2 = t2
W [y1 , y2 ]
t
v10 =
So,
Z
1
−t3 dt = − t4
4
Z
1
v2 = t2 dt = t3
3
v1 =
Then,
yp (t) = y1 v1 + y2 v2
−t4
t3
+ t2
4
3
t5
=
12
=t
(c)
y(t) = c1 y1 (t) + c2 y2 (t) + yp (t)
= c1 t + c2 t2 +
t5
.
12
Problem 2: (15 points) True/False. If the statement is always true, mark true. Otherwise, mark
false. You do not need to show your work. (Any work will not be graded.)
(a) There exists a real 2 × 2 matrix, A, with eigenvalues λ1 = 1, λ2 = i.
(b) If λ = 2 is a repeated eigenvalue of multiplicity 2 for a matrix A, then the eigenspace
associated with λ has dimension 2.
(c) If an n × n matrix has n distinct eigenvalues, then each eigenvalue has a corresponding
one-dimensional eigenspace.
Solution:
(a) False. Complex eigenvalues of real
are always present in complex conjugate pairs.
matrices
2 1
(b) False. Not necessarily, e.g. A =
.
0 2
(c) True. If {λ1 , . . . , λn } is a set of distinct eigenvalues then a corresponding set of eigenvectors
{v1 , . . . , vn } is linearly independent. This is a maximally linearly independent set in Rn
so if any one eigenvalue had two (or more) linearly independent eigenvectors, then one of
them would be a linear combination of eigenvectors of other eigenspaces, which would be
a contradiction.
Problem 3: (20 points) Consider the following differential equation,
x00 + 2x0 + x = et
(a) Find the general solution x(t).
(b) Solve the initial value problem with x(0) = x0 (0) = 0.
Solution:
(a) Homogeneous solution:
x00h + 2x0h + xh = 0
⇒ xh = c1 e−t + c2 te−t
Undetermined coefficients:
xp = Aet
⇔
⇔
(Aet )00 + 2(Aet )0 + (Aet ) = et
1
A=
4
General Solution:
x(t) = xh (t) + xp (t)
1
= c1 e−t + c2 te−t + et .
4
(b)
1
1
⇒ c1 = −
4
4
1
1
x0 (0) = 0 ⇒ 0 = −c1 + c2 + ⇒ c2 = −
4
2
x(0) = 0 ⇒ 0 = c1 +
So,
1
1
1
x(t) = − e−t − te−t + et .
4
2
4
Problem 4: (20 points) Find the general solution to the system of differential equations
1 2
0
~x =
~x.
−2 1
Solution: p(λ) = (1 − λ)2 + 4 = λ2 − 2λ + 5 ⇒ λ = 1 ± 2i.
Find the eigenvector corresponding to λ = 1 + 2i: Solve (A − λI)~v = ~0:
−2i 2 0
−2i 2 0
∼
−2 −2i 0
0 0 0
1
1
0
⇒ ~v =
=
+i
.
i
0
1
So,
~xRe
~xIm
cos 2t
= e (cos βt p~ − sin βt ~q) = e
− sin 2t
αt
t sin 2t
.
= e (sin βt p~ + cos βt ~q) = e
cos 2t
αt
t
Then the general solution is,
~x(t) = c1 ~xRe + c2 ~xIm
c1 cos 2t + c2 sin 2t
t
=e
−c1 sin 2t + c2 cos 2t
Problem 5: (25 points) Consider the matrix


1 0 −1
A = 1 2 0  .
0 0 2
(a) Find all the eigenvalues of A.
(b) For each eigenvalue that you found in part (a), find a basis for the corresponding eigenspace.
(c) Find the general solution of ~x0 = A~x. (Hint: you need to find a generalized eigenvector)
Solution:
(a) The characteristic polynomial of A is p(λ) = (1 − λ)(2 − λ)2 . The eigenvalues are λ1 =
1, λ2 = 2 (repeated).
(b)
λ1 = 1: Solve (A − λ1 I)~v1 = ~0:


 
−1
0 0 −1 0
1 1 0 0 ⇒ ~v1 =  1  .
0 0 1 0
0
Then ~v1 is a basis for the eigenspace E1 corresponding to λ1 = 1.
λ2 = 2: Solve (A − λ2 I)~v2 = ~0:


 
−1 0 −1 0
0
 1 0 0 0 ⇒ ~v2 = 1
0 0 0 0
0
So ~v2 is a basis for the eigenspace E2 corresponding to λ2 = 2.
(c) ~x0 = A~x is a 3 × 3 system, so we need three linearly independent solutions. Two solutions
are given by ~x1 (t) = eλ1 t~v1 and ~x2 (t) = eλ2 t~v2 . For the third, we need to find a generalized
eigenvector for the eigenvalue λ2 = 2, since this is the repeated eigenvalue.
Solve (A − λ2 I)~u = ~v2 :


   
−1 0 −1 0
0
1
 1 0 0 1 ⇒ ~u = s 1 +  0 
0 0 0 0
0
−1
for a free parameter s. Since we only need one such vector, set s = 0. Then a third linearly
independent solution is given by
~x3 (t) = teλ2 t~v2 + eλ2 t ~u
 
 
0
1
= te2t 1 + e2t  0  .
0
−1
The general solution is given by
~x(t) = c1 ~x1 (t) + c2 ~x2 (t) + c3 ~x3 (t)
 
 
 
−1
0
1
= c1 et  1  + c2 e2t 1 + c3 e2t  t 
0
0
−1


−c1 et + c3 e2t
t

= c1 e + c2 e2t + c3 te2t  .
−c3 e2t