MATH 2Z03 —SAMPLE 1

Duration of Examination: 3 hours

M2Z03: SAMPLE FINAL EXAM C

MATH 2Z03 —SAMPLE 2

6 x2
1. The general solution of the differential equation y 0 = is
2 y + cos y
(A) y 2 + sin y = 2 x3 + c (D) y 2 + cos y = x3 + c
(B) 2 y 2 + cos y = 6 x3 + c (E) y 2 + sin y = x2 + c

(C) 2 y 2 + sin y = 6 x3 + c

2. The solution of the initial value problem (1 + x2 )y 0 + 2x y = cos x, y(0) = 1 is

1 + x2 1 + sin(x)
(A) y = (D) y =
sin(x) x2 + 1
x2 + 2 2 − cos(x)
(B) y = (E) y =
2 + sin(x) x2 + 1

x2 + 1
(C) y =
1 + cos(x)

√
3. Which of the following homogeneous linear differential equations has y(x) = x sin( 2 x)
as a solution?

(A) y 00 + 2 y = 0 (D) y (4) + 2 y 00 + 2 y = 0

(B) y 00 + 2 y 0 + 2 y = 0 (E) None of the above

(C) y (4) + 4 y 00 + 4 y = 0

4. Which of the following is a fundamental set of solutions for the homogeneous differential
equation y 00 − 6 y 0 + 9 y = 0 ?

(A) {e3x , e3x } (D) {e−3x , x e−3x }

(B) {e−3x , e3x } (E) {ex/3 , e−x/3 }

(C) {e3x , x e3x }

5. The Wronskian determinant W (x) associated with the set of functions {x, ex , x + ex } is

(A) ex (D) 0
(B) e2x (E) 1
(C) xe2x
MATH 2Z03 —SAMPLE 3

6. The homogeneous equation y 00 + 16y = 0 has for general solution

y(x) = c1 cos(4x) + c2 sin(4x)

(you need not check that). The boundary–value problem y 00 +16 y = 0, y(0) = 0, y(π) = 1
has:

(A) the unique solution y = 1 − cos(4x) (D) two solutions: y = cos(4x) and y = sin(4x)
(B) no solution (E) infinitely many solutions: y = C sin(4x),
where C is any number
(C) the unique solution y = sin(4x)

7. The homogeneous differential equation y (4) + y (3) − 2 y 00 = 0 has for general solution

y(x) = c1 + c2 x + c3 e−2 x + c4 ex

(you need not check that). For this equation, the initial value problem y(0) = 1,
y 0 (0) = −1, y 00 (0) = 4, y (3) (0) = −8 has solution

(A) y(x) = 0 (D) y(x) = x + e−2 x

(B) y(x) = x + e−x (E) y(x) = 1 + e−x
(C) y(x) = 1 + e−2 x

x
8. The solution of the initial value problem 4 y 00 − y = x e 2 subject to initial condition
y(0) = 1, y 0 (0) = 0 is
3 x 1 x 1 x 1 x
(A) y(x) = e 2 + e− 2 + x2 e 2 − xe 2
4 4 8 4
x x x x
(B) y(x) = e 2 + e− 2 + x2 e 2 − xe 2
3 −x 1 x 1 x 1 x
(C) y(x) = e 2 + e 2 + x2 e− 2 − xe− 2
4 4 8 4
7
(D) y(x) =
8
(E) None of the above
 MATH 2Z03 —SAMPLE 4

9. If the method of undetermined coefficients is used to solve the linear differential equation
y 00 + 9 y = sin(3 x), then the form of a particular solution is

(A) yp (x) = A sin(3 x) + B cos(3 x)

(B) yp (x) = A x sin(3 x) + B x cos(3 x)

(C) yp (x) = A sin(3 x)
(D) yp (x) = A x2 sin(3 x) + B x2 cos(3 x)

(E) yp (x) = A x3 sin(3 x) + B x3 cos(3 x)

10. The general solution of the linear differential equation x2 y 00 − 6 x y 0 − 8 y = x2 , for which
the homogeneous part has the solutions y1 (x) = x8 and y2 (x) = x1 , is given by
c2 1
(A) y(x) = c1 x8 + −
x 48 x6
c2 1
(B) y(x) = c1 x8 + −
x 27 x3
c2 1
(C) y(x) = c1 x8 + −
x 48 x2
c2 1 4
(D) y(x) = c1 x8 + − x
x 12
c2 1 2
(E) y(x) = c1 x8 + − x
x 18

11. Consider the boundary–value problem y 00 + λ y = 0, y 0 (0) = 0, y 0 (π) = 0. The smallest

eigenvalue λ for this problem is

1
(A) 1 (D) 4
1
(B) 2 (E) π
(C) 0
 MATH 2Z03 —SAMPLE 5

12. Suppose that A is a 2 × 2 real matrix and that one of its eigenvalue is λ = 2 + i with
2
corresponding eigenvector v = . Then, the general solution of the 1st order system
i
X0 = AX is given by
 t   t   t   2t 
e cos(2t) e sin(2t) 2e cos t e sin t
(A) X(t) = c1 + c2 (D) X(t) = c 1 + c 2
−2et sin(2t) 2et cos(2t) −et sin t 2e2t cos t
 2t   2t   2t   2t 
2e cos t 2e sin t 2e sin t 2e cos t
(B) X(t) = c1 + c2 (E) X(t) = c 1 + c 2
−e2t sin t e2t cos t −e2t sin t e2t cos t
 2t   2t 
2e sin t −e sin t
(C) X(t) = c1 + c2
−e2t cos t 2e2t cos t

13. Suppose that A is a 3 × 3 real matrix with characteristic polynomial

P (λ) = det(A − λI) = −λ3 + 2λ + 4.

Find A if  
8 4 −2
A3 = 0 2 2 .
0 −2 2
   
1 1 −2 0 2 −1
(A) A =  2 1 −2 (D) A = 2 1 2
−2 −2 2 4 −1 2
   
2 1 −2 2 2 −1
(B) A = 1 2 −1 (E) A = 0 −1 1
0 −2 1 0 −1 −1
 
0 1 −2
(C) A = 2 1 1
1 −2 4
 MATH 2Z03 —SAMPLE 6

14. The symmetric matrix  

8 −2 −2
A = −2 11 −1 .
−2 −1 11
 
1
has characteristic polynomial P (λ) = −(λ − 6) (λ − 12)2 and v1 = −3 is an eigen-
1
vector associated with the eigenvalue λ = 12. Another eigenvector associated with the
eigenvalue λ = 12 and orthogonal to v1 is given by
   
−4 3
(A) v2 =  1  (D) v2 = 1
7 0
   
−1 1
(B) v2 =  0  (E) v2 =  2 
1 −4
 
2
(C) v2 = −1
−1

15. Let  
2 a
A=
3 −1
Find the value of a for which λ = 5 is an eigenvalue of A.
(A) a = 2 (D) a = 6

(B) a = −5 (E) a = 7
(C) a = −3

 
1
16. Find the value of a for which the vector v = 2 is an eigenvector of the matrix
2
 
−1 0 2
A= 0 1 a .
a 2 0

(A) a = 1 (D) a = 2
(B) a = −1 (E) a = 0
(C) a = −3
 MATH 2Z03 —SAMPLE 7

17. The eigenvalues λ of the matrix

 
2 0 1
A =  0 2 2
13 1 0

are given by:

(A) λ = 0, 2, 3 (D) λ = −1, 1, 2

(B) λ = −2, 1, 4 (E) λ = −13, 2, 3

(C) λ = −3, 2, 5

18. A 3 × 3 matrix A has eigenvalues λ1 = −2, λ2 = 1 and λ3 = 4 with corresponding

eigenvectors      
−1 1 0
v1 =  0  , v2 = 1 , v3 = 1 ,
2 1 2
respectively. Then, a matrix P satisfying
 
4 0 0
P −1 AP = 0 −2 0
0 0 1

is given by:
   
−1 0 2 1 0 −1
(A) P =  1 1 1 (D) P = 1 1 0 
0 1 2 1 2 2
   
−1 1 0 0 −1 1
(B) P =  0 1 1 (E) P = 1 0 1
2 1 2 2 2 1
 
1 −1 0
(C) P = 1 0 1
1 2 2
 MATH 2Z03 —SAMPLE 8

19. The 3rd order linear differential equation x000 (t) − 2 x00 (t) + 3 x(t) = 0 can
 be written
x(t)
as a 1st order linear system of the form X0 (t) = A X(t) where X(t) = y(t), with
z(t)
y(t) = x0 (t), z(t) = x00 (t) and A being the matrix
   
1 0 0 0 1 0
(A) 0 1 0  (D) 1 0 1
0 3 −2 3 −2 1
   
0 1 0 1 1 1
(B) 1 0 0 (E)  1 1 0
2 0 3 −3 2 0
 
0 1 0
(C)  0 0 1
−3 0 2

20. The solution of the initial value problem

(
x0 (t) = −x(t),
y 0 (t) = −4 x(t) + 3 y(t),

with initial conditions x(0) = 2, y(0) = 3 is given by

(A) x(t) = e−t + e3t , y(t) = e−t + 2 e3t (D) x(t) = −e−t + 2 e3t , y(t) = −e−t + 3 e3t

(B) x(t) = 3 e−t − e3t , y(t) = 2 e−t + e3t (E) x(t) = 2 e−t , y(t) = −e−t + 4 e3t
(C) x(t) = 2 e−t , y(t) = 2 e−t + e3t

21. Suppose that A is a real, 3 × 3 matrix with eigenvalues λ1 = −1, λ2 = 0 and λ3 = 2 and
corresponding eigenvectors
     
1 1 0
v1 = 0 , v2 = 1 , v3 = 1 ,
2 1 2
 
x(t)
respectively. Let X(t) = y(t) be the unique solution of the initial value problem
z(t)  
3
X0 (t) = A X(t) with initial condition X(0) = 3. Then, the value of the 1st component
9
of X(t) at t = 1 is
(A) x(1) = 2 + e (D) x(1) = −1 + e−1 + 2 e
(B) x(1) = e + 2 e−1 (E) x(1) = 1 − 2 e−1 − e
(C) x(1) = 1 + 2 e−1
 MATH 2Z03 —SAMPLE 9

22. Let A be an n × n real, symmetric matrix. Which of the following statements is not
always correct?
(A) A B = B A for any n × n symmetric matrix B
(B) A is diagonalizable
(C) A2 − A is symmetric
(D) A2 is symmetric
(E) A2 is diagonalizable

23. Let A be an n × n real matrix. Which of the following does not necessarily imply that
A is diagonalizable:
(A) A has n distinct eigenvalues.
(B) A is symmetric.
(C) A = B 2 , where B is diagonalizable.
(D) A is invertible and A−1 is diagonalizable.
(E) The 1st order system X0 (t) = A X(t), where X(t) is a column vector with n com-
ponents, admits n linearly independent solutions for −∞ < t < ∞.

24. The Laplace transform L (3 t − 1)2 is



18 6 1 (3 s − 1)2
(A) − 2+ (D)
s3 s s s3
9 6 1 s/3
(B) − +1 (E) se
s2 s 9
(C) 9 s e−s/3

25. The Laplace transform L e3t−2 is



1 e−2
(A) (D)
3s − 5 s−3
3 1
(B) 2 (E)
s− 3
(3 s − 2)3

e2
(C)
s+3

26. The Laplace transform L {t sin(2t)} is

4 2s
(A) 2 (D)
(s + 4)2 (s2 + 4)2
4s 2 s2
(B) (E) −
(s2 + 4)2 s2 +4
4s
(C) −
(s2 + 4)2