Download Exam 3 Sol

APPM 2360: Section exam 3
7:00pm – 8:30pm, November 28, 2012.
ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number,
(3) recitation section (4) your instructor’s name, and (5) a grading table. Text books, class notes,
and calculators are NOT permitted. A one-page crib sheet is allowed.


1 0 1
Problem 1: (20 points) Given the matrix A =  0 1 2 ,
−2 0 3
(a) (14 points) Find the eigenvalues and eigenvectors of matrix A.
(b) (6 points) Find the eigenspaces for matrix A and give their dimension.
Solution:
(a) det(A − λI)v = (λ − 1)(−λ2 + 4λ − 5) = 0 leads to the eigenvalues λ1 = 2 + ı, λ2 = 2 − ı
~0 yields the corresponding eigenvectors ~v1 =
and
det(A
− λI)v

=
 1 λ3 ı = 1, and solving
1
ı
0
2 − 2
2 + 2
 1 − ı , ~v2 =  1 + ı  and ~v3 =  1 .
0
1
1
(b) Eλ1 = span {~v1 } with dim(Eλ1 ) = 1
Eλ2 = span {~v2 } with dim(Eλ2 ) = 1
Eλ3 = span {~v3 } with dim(Eλ3 ) = 1
Problem 2: (20 points) Consider the second order differential equations
5y 00 − 10 y 0 + 15 y = 0
(a) (10 points) Construct the general solution.
(b) (4 points) Construct the particular solution with initial condition y(0) = −2, y 0 (0) = 1.
(c) (6 points) Letting x1 (t) = y(t) and x2 (t) = y 0 (t), construct the linear system
of
two first
x1
0
order differential equations for x1 (t) and x2 (t), i.e., ~x = A~x where ~x =
. (Do not
x2
solve the linear system.)
Solution:
√
(a) Characteristic equation: r2 − 2r +√
3 = 0. A pair√of complex conjugate roots r1,2 = 1 ± 2.
General solution: y(t) = et (c1 cos( 2t) + c2 sin( 2t)).
√
√
√
√
(b) Note that y 0 (t) = et ((c1 + 2c2 ) cos( 2t)+(c2 − 2c1 ) sin( 2t)). Applying initial condition,
a linear system is obtained:
c1 +
So c1 = −2 and c2 =
√3 .
2
√
c1 = −2
2c2 = 1.
√
The particular solution is y(t) = et (−2 cos( 2t) +
(c)
x01 = x2
x02 = −3x1 + 2x2 .
√3
2
√
sin( 2t)).
Problem 3: (20 points)
(a) (12 points) Use the Method of Undetermined Coefficients to write down the general form
of the particular solution, yp , for the differential equations given below but do not solve:
(i) y 00 + 6y 0 + 9y = e−3t
(ii) y 00 + 6y 0 + 9y = t2 sin(6t)
(iii) y 00 + 6y 0 + 9y = te2t cos(t)
(iv) y 00 + 9y = −6y 0 + e−2t + t2 e−2t
(b) (8 points) Now use the Method of Undetermined Coefficients to find a particular solution
of the equation,
y 00 + 6y 0 + 9y = e−3t
Solution:
(a)(i) yp = At2 e−3t , since y = e−t and y = te−3t are solutions of the corresponding homogeneous
equation.
(ii) yp = (At2 + Bt + C) sin(6t) + (Dt2 + Et + F ) cos(6t)
(iii) yp = (At + B)e2t sin(t) + (Ct + D)e2t cos(t)
(iii) yp = (At2 + Bt + C)e−2t
(b) If yp = At2 e−3t , then yp0 = 2Ate−3t − 3At2 e−3t , and yp00 = 2Ae−3t − 12Ate−3t + 9Ate−3t , now
1
substituting yp into the equation y 00 + 6y 0 + 9y = e−3t yields A = 1/2 and so yp = − t2 e−3t .
2
PLEASE TURN OVER
Problem 4: (20 points) Consider the following nonhomogeneous Euler equation with initial
conditions
t2 x00 + 3tx0 − 3x = t
x(1) = 0,
x0 (1) = 0
(a) (6 points) Find the general homogenous solution by assuming the form xh (t) = tr .
(b) (9 points) Find the general solution to the nonhomogenous problem.
(c) (5 points) Find the solution to the initial-value problem.
Solution:
(a) The characteristic equation is given by r(r − 1) + 3r − 3 = r2 + 2r − 3 = 0. Therefore
r1 = 1, r2 = −3. The general homogenous solution is given by
xh (t) = c1 t + c2 t−3
(b) Solve the normalize problem
3
3
1
x00 + x0 − 2 x =
t
t
t
by variation of parameters where xp (t) = v1 (t)t + v2 (t)t−3 . Therefore
0 0
t
t−3
v1
=
v20
et
1 −3t−4
The Wronksian for the homogenous solutions is −4t−3 . Using Cramer’s rule:
0
t 0 t−3 −1
t
1 t−1 −3t−4 1
t3
0
0
v1 =
=
,
v
=
=
−
2
−4t−3
4t
−4t−3
4
Therefore
1
1
v1 (t) = ln(t),
v2 (t) = − t4
4
16
Hence
1
1
1
1
xp (t) = ln(t)t − t4 t−3 = t ln(t) − t
4
16
4
16
The general solution is thus given by
t
xg (t) = c1 t + c2 t−3 +
(4 ln(t) − 1)
16
(c) Solve the initial value problem gives
1 1 1
c1
16
=
3
1 −3
c2
− 16
Therefore
1
1 1
16
3
1 −3 − 16
1
1 1
1 1
16
∼
∼
0 −4 − 14
0 1
1 0 0
∼
1
0 1 16
Hence
x(t) =
1
16
1
16
1 −3
t
t +
(4 ln(t) − 1)
16
16
Problem 5: (20 points) Give a short answer to the following statements.
∼
(a) (4 points) Let A be a 4 × 4 real matrix. Two distinct complex eigenvalues λ1 and λ2 of A
are known. Compute the remaining eigenvalues.
(b) (4 points) Let A be a n×n real matrix. If λ1 is an eigenvalue of A with algebraic multiplicity
2, then rank (A − λ1 I) = n − 2. (TRUE/FALSE)
(c) (4 points) Any solution y(t) to y 00 + 5y 0 + 6y = 0 tends to zero as t → ∞. (TRUE/FALSE)
(d) (4 points) Let A be a 3 × 3 real matrix with real eigenvalues. The eigenvectors of A form
a basis for R3 . (TRUE/FALSE)
(e) (4 points) Let A be a n × n real matrix with λ1 one of its eigenvalues. Then (A − λ1 I) is
an invertible matrix. (TRUE/FALSE)
Solution:
(a) λ3 = λ̄1 and λ4 = λ̄2
(b) FALSE
(c) TRUE (Roots of the characteristic equation are −2 amd −3)
(d) FALSE (Linearly independent eigenvectors might not be obtained for repeated eigenvalues)
(e) FALSE (det (A − λ1 I) has to be zero to obtain a non-trivial eigenvector)