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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
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
```
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