Differential Equations and Linear Algebra - Fall 2017

Transcription

1 NAME: Differential Equations and Linear Algebra - Fall 27 Final Exam, December 4, 27 GRADING:. In multiple choice problems -3 you don t have to show your work. Consequently, no partial credit will be given. 2. In multiple choice problems 4-8: (a) If your answer is correct and you show all important steps, you ll get full credit for this problem. (b) If your answer is correct, but you show only few important steps, you ll get 9% for this problem. (c) If your answer is correct, but you show no or little or irrelevant work, you ll get 8% for this problem. (d) If your answer is incorrect, you ll get for this problem. Average 7.9% Max % (2 students) Min 38% (3 students)

2 . Let W be the set of singular 2 2 matrices under the usual operations. Then A W is not a vector space, since it is not closed under addition. correct B W is not a vector space, since is not in W. C W is not a vector space, since it is not closed under multiplication by a scalar. D W is not a vector space, since it is not a subspace of the vector space of 2 2 matrices. Info: The average in the class for this problem was 34.3%. 2. Let Find A. [ ] 3 π A πe 3 2 e 2 [ ] 2 e B 3 2 πe π 3 [ ] 2 e C πe 3 2 π 3 [ ] 3 π D 3 2 πe e 2 [ 2 π A = e 3 correct ] Info: The average in the class for this problem was 77.5%. 3. Vectors v,...,v p are said to be linearly dependent if A there exist scalars c,...,c p, such that c v +...+c p v p =. B there exist scalars c,...,c p, not all zero, such that c v +...+c p v p. C there exist scalars c,...,c p, such that c v +...+c p v p. D the vector equation c v +...+c p v p = has only the trivial solution. E there exist scalars c,...,c p, not all zero, such that c v c p v p =. correct F the vector equation c v +...+c p v p = has the trivial solution. G None of the above Info: The average in the class for this problem was 65.7%.

3 4. Solve (+y 2 ) dy dx = xcosx. A y 3 +3y = 3(xsinx cosx)+c B y 3 +3y = 3(xsinx+cosx)+c correct C y 3 +3y = 3(xsinx+cosx)+c D y 3 +3y = 3(xsinx cosx)+c Info: The average in the class for this problem was 95.%. 5. Solve dy dx y = e2x. A y = e x (e 2x +c) B y = e 2x (e x +c) C y = e x (e x +c) correct D y = e 2x (e 2x +c) Info: The average in the class for this problem was 96.%. 2

4 6. Solve [+ln(xy)]dx+xy dy =. A yln(xy) = c B xyln(xy) = c C xlny = c D ylnx = c E x ln(xy) = c correct F xylnx = c G xylny = c H None of the above Info: The average in the class for this problem was 49%. 3

5 7. Solve ydx (2x+y 4 )dy =. A xy 2 +y 2 /2 = c B xy 2 y 2 /2 = c correct C x 2 y y 2 /2 = c D x 2 y +y 2 /2 = c Info: The average in the class for this problem was 55.9%. 4

6 8. Compute the second Picard iteration for the initial-value problem dy dx A y 2 (x) = 3+4x 2 +6x 4 B y 2 (x) = 3+4x 2 6x 4 C y 2 (x) = 3 4x 2 +6x 4 D y 2 (x) = 3 4x 2 6x 4 correct y 2 (x) = 3+6x 2 +6x 4 = 4xy, y() = 3. Info: The average in the class for this problem was 47.%. 9. Solve y y 2y =. A y = c e x +c 2 xe 2x B y = c e x +c 2 xe 2x C y = c e x +c 2 xe 2x D y = c e x +c 2 xe 2x correct Info: The average in the class for this problem was 72.6%. 5

7 . Find a solution of the equation y +2y +y = 5sin3t. A y = 4sin3t+3cos3t B y = 4sin3t 3cos3t C y = 4sin3t 3cos3t correct D y = 4sin3t+3cos3t Solution: We first note that the right-hand side of y +2y +y = 5sin3t is the imaginary part of 5e i3t = 5cos3t+5isin3t Therefore we will find a particular solution ψ(t) of the given equation as the imaginary part of a complex-valued solution φ(t) of the equation L[y] = d2 y dt 2 +2dy dt +y = 5e3it () To this end, observe that α = 3i is not a root of the the characteristic equation r 2 +2r + = Therefore we set We have φ(t) = A e 3it (2) φ = (A e 3it ) = A (e 3it ) = A e 3it (3it) = A e 3it (3i) = 3A e 3it i and therefore φ = (3A e 3it i) = 3A i(e 3it ) = 3A ie 3it (3it) = 3A ie 3it (3i) = 9A i 2 e 3it = 9A e 3it Using this in () gives L[φ](t) = φ +2φ +φ = 9A e 3it +6A e 3it i+a e 3it = 8A e 3it +6A e 3it i = A e 3it ( 8+6i) A e 3it ( 8+6i) = 5e 3it A ( 8+6i) = 5 A = i

8 Note that A can be rewritten as since A = 5 8+6i = A = 4 3i 5( 8 6i) ( 8+6i)( 8 6i) = 5( 8 6i) ( 8) 2 (6i) = 5( 8 6i) = 5( 8 6i) = 4 3i Substituting this into (2), we get φ(t) = A e 3it = ( 4 3i)e 3it = ( 4 3i)(cos3t+isin3t) = ( 4)cos3t+( 4)isin3t+( 3i)cos3t+( 3i 2 )sin3t = ( 4)cos3t+( 4)isin3t+( 3i)cos3t+3sin3t = ( 4cos3t+3sin3t)+i( 4sin3t 3cos3t) Therefore ψ(t) = Im { φ(t) } = 4sin3t 3cos3t is a particular solution of y +2y +y = 5sin3t. Info: The average in the class for this problem was 56.9%. 7

9 . Find a solution of the equation y +xy = (3n )( ) n A y(x) = x+ (3n+)! n= (2n )( ) n B y(x) = x+ (2n+)! n= (3n )( ) n C y(x) = x+ (3n+)! n= (2n )( ) n D y(x) = x+ (2n+)! n= x 3n x 2n+ x 3n+ x 2n correct Solution: We set Computing and y(x) = a +a x+a 2 x 2 +a 3 x 3 +a 4 x = y = a +2a 2 x+3a 3 x 2 +4a 4 x = y = 2a 2 +6a 3 x+2a 4 x = we see that y(x) is a solution of y +xy = if y +xy = n(n )a n x n 2 +x a n x n = a n x n na n x n n(n )a n x n 2 n(n )a n x n 2 + a n x n+ = () Our next step is to rewrite the first summation in () so that the exponent of the general term is n+, instead of n 2. We have n(n )a n x n 2 = (n+3)(n+2)a n+3 x n+ n= 3 = ( 3+3)( 3+2)a 3+3 x 3+ +( 2+3)( 2+2)a 2+3 x 2+ +( +3)( +2)a +3 x + + (n+3)(n+2)a n+3 x n+ = ++2a 2 + (n+3)(n+2)a n+3 x n+ 8

10 therefore we can rewrite () in the form 2a 2 + (n+3)(n+2)a n+3 x n+ + Therefore the coefficients must satisfy That is, a n x n+ = 2a 2 + [(n+3)(n+2)a n+3 +a n ]x n+ = a 2 = and (n+3)(n+2)a n+3 +a n = a n a 2 = and a n+3 = (n+3)(n+2) To find a solution of y +xy =, we set a =, a =. In this case, a,a 3,a 6,... and a 2,a 5,a 8,... are zero since a =, a 3 = a 3 2 = 3 2 =, a 6 = a = 6 5 = and a 2 =, a 5 = a = 5 4 =, a 8 = a = 8 7 = and so on. The coefficients a,a 4,a 7,... are determined from the relations a = a 4 = a 4 3 = 4 3 = = 2 4! a 7 = a = 2 4! 7 6 = 2 5 4! = 2 5 7! a = a 7 9 = 2 5 7! 9 = ! 9 8 = 2 5 8! a 3 = a 3 2 = 2 5 8! 3 2 = 2 5 8! 3 2 = ! and so on. Proceeding inductively, we find that Hence, is a solution of y +xy =. y 2 (x) = x 2 4! x ! = x+ n= a 3n+ = ( ) n (3n ) (3n+)! x ! ( ) n (3n ) (3n+)! Info: The average in the class for this problem was 32.4%. 9 x x ! x 3n+

11 2. Find a basis of the vector space of all solutions of the system x +x 2 +x 3 +x 4 +x 5 +x 6 = x 3 +x 4 +x 5 +x 6 = x 4 +x 5 +x 6 = A,, B,, C,, correct D,, Info: The average in the class for this problem was 58.8%.

12 3. Find the general solution of the system ẋ = Ax, where A = Hint: The eigenvalues of A are λ = and λ 2 = 2. A x(t) = c e t B x(t) = c e t +c 2 e 2t C x(t) = c e t +c 2 e 2t +c 2 e 2t +c 3 e 2t D x(t) = c e t +c 2 e 2t 3 2 +c 3 e 2t +c 3 e 2t +c 3 e 2t correct Info: The average in the class for this problem was 66.7%.

13 4. Find the general solution of the system ẋ = Ax, where [ ] 2 A = 2 [ ] [ ] sin2t cos2t A x(t) = c +c 2 cos2t sin2t [ ] [ ] sin2t cos2t B x(t) = c +c 2 cos2t sin2t [ ] [ ] sin2t cos2t C x(t) = c +c 2 cos2t sin2t [ ] [ ] sin2t cos2t D x(t) = c +c 2 correct cos2t sin2t Info: The average in the class for this problem was 52.9%. 2

14 5. Find the general solution of the system ẋ = Ax, where Hint: The eigenvalue of A is λ = 4. A x(t) = c e 4t 3 +c 2 e 4t A = t +t B x(t) = c e 4t +c 2 e 4t t C x(t) = c e 4t D x(t) = c e 4t 3 3 +c 2 e 4t 3 3t t 3t t +c 2 e 4t t 3t c 3 e 4t +c 3 e 4t +t t 3t +c 3 e 4t c 3 e 4t correct 2 Info: The average in the class for this problem was 7.6%. 3

15 6. Let A = [ ] [ ] +4t 8t A e 2t 2t 4t [ ] +4t 8t B e 2t 2t 4t [ ] 4t 8t C e 2t 2t 4t [ ] +4t 8t D e 2t 2t +4t, then e At is correct Info: The average in the class for this problem was 62.8%. 4

16 7. Determine whether each solution x(t) of the differential equation [ ] ẋ = x 2 is stable or unstable. A Unstable, because all the eigenvalues of A have negative real part. B Stable, because at least one eigenvalue of A has positive real part. C Unstable, because at least one eigenvalue of A has positive real part. correct D Stable, because all the eigenvalues of A have negative real part. Info: The average in the class for this problem was 9.2%. 5

17 8. Expand the function in a pure sine series. Then where A b n = 4 nπ B b n = 4 nπ C b n = 4 nπ D b n = 4 nπ ( sin 2nπ ( cos 2nπ 3 ( )n 3 ( )n (cos nπ ) 3 ( )n (sin nπ 3 ( )n ) f(x) = ) ) f(x) = { for < x 2 2 for 2 < x < 3 n= b n sin nπx 3 correct Info: The average in the class for this problem was 67.7%. 6