Mathematics 38 Differential Equations

Exam III December 5, 2011

No calculators, books or notes are allowed on the exam. All electronic devices must be turned off

and put away.You must show all your work in the blue book in order to receive full credit. Please

box your answers and cross out any work you do not want graded. Make sure to sign your blue

book. With your signature you are pledging that you have neither given nor received assistance on

the exam.

Good luck!

1. (10 points) Given the matrix A=1−3

−3 1 and the eigenvector ~v =1

1ﬁnd

a. the eigenvalue λof Ato which ~v corresponds and

b. the associated solution of D~x =A~x.

Solution: a. 1−3

−3 1 1

1=−2

−2=−21

1so λ=−2.b. e−2t1

1.

2. (20 points) The 3×3matrix A=

3−3 1

100

010

has a triple eigenvalue of 1(you do not have to

verify this). Find the general solution of

D~x =A~x.

Solution: ~v1=

1

0

0

,~v2=

0

1

0

,~v3=

0

0

1

are (obviously!!) generalized eigenvectors.

Each of these gives a solution hi(t) = et[I+t(A−I) + t2

2(A−I)2]~vi=

=et

1 0 0

0 1 0

0 0 1

+t

2−3 1

1−1 0

0 1 −1

+t2

2

1−2 1

1−2 1

1−2 1

~vi

=et

1 + 2t+t2/2−3t−t2t+t2/2

t+t2/2 1 −t−t2t2/2

t2/2t−t21−t+t2/2

~vi,

so the sought general solution is

~x(t) = c1et

1 + 2t+t2/2

t+t2/2

t2/2

+c2et

−3t−t2

1−t−t2

t−t2

+c3et

t+t2/2

t2/2

1−t+t2/2

3. (20 pts) Find the general solution of ~x′=1−1

1 1 ~x. Hint: 1

−iis an eigenvector.

Solution: 1−1

1 1 1

−i=1 + i

∗= (1+i)1

∗, so the eigenvector 1

−icorresponds

to the eigenvalue 1 + i, and the corresponding (complex) solution is

e(1+i)t1

−i=et(cos t+isin t)1

−i=etcos t+isin t

sin t−icos t=etcos t

sin t+ietsin t

−cos t.

The general solution then is ~x(t) = c1etcos t

sin t+c2etsin t

−cos t.

1

## Document Summary

No calculators, books or notes are allowed on the exam. All electronic devices must be turned off and put away. you must show all your work in the blue book in order to receive full credit. Please box your answers and cross out any work you do not want graded. With your signature you are pledging that you have neither given nor received assistance on the exam. Good luck: (10 points) given the matrix a = (cid:18) 1 3. 1(cid:19) and the eigenvector ~v = (cid:18) 1. 3 a . the eigenvalue of a to which ~v corresponds and: the associated solution of d~x = a~x. Solution: a. (cid:18) 1 3: (20 points) the 3 3 matrix a = . 1(cid:19) so = 2. b. e 2t(cid:18) 1. Has a triple eigenvalue of 1 (you do not have to. 1 2 so the sought general solution is. Each of these gives a solution hi(t) = et[i + t(a i) + t.