Set of Functions

Solutions. Signature: Do not write your student ID number on this front page. Please write your student ID number in the space provided on the second page. Take your time to read the entire paper before you begin to write, and read each question carefully. Remember that certain questions are worth more points than others. Make a note of the questions that you feel confident you can do, and then do those first: you do not have to proceed through the paper in the order given. • You have 75 minutes to complete this exam. • This is a closed book exam, and no notes of any kind are allowed. The use of cell phones, pagers or any text storage or communication device is not permitted. • Only the Faculty approved TI-30 calculator is allowed. • The correct answer requires justification written legibly and logically: you must convince me that you know why your solution is correct. Answer these questions in the space provided. Use the backs of pages if necessary. • Where it is possible to check your work, do so. • Good Luck! 1 Student number: Problem 1 2 3 4 5 6 7 8 Total Points 6 16 21 10 25 20 22 21 BONUS 120 Your Marks Question 1. [6 points] Determine whether the given set of fuctions is linearly independent on the interval (−∞, ∞). (a) [3 points] f 1 (x) = x, f 2 (x) = x − 1, f 3 (x) = x + 3 Solution: Since (−4)x + 3(x − 1) + 1(x + 3) = 0 the set of functions is linearly dependent. (b) [3 points] f 1 (x) = 2 + x, f 2 (x) = 2 + |x| Solution: From the graphs of f 1 (x) = 2 + x and f 2 (x) = 2 + |x| we see that the set of functions is linearly independent since they cannot be multiples of each other. The graph of f 2 (x) = 2 + |x| is The graph of f 1 (x) = 2 + x is 2 Question 2. [16 points] For the differential equation 4y 00 − 4y 0 + y = 0, (a) [14 points] verify the fuctions of y 1 = e x/2 , y 2 = xe x/2 form a fundamental set of solutions for the above differential equaiton on the interval of (−∞, ∞); Solution: • Check that y 1 is a solution of 4y 00 − 4y 0 + y = 0. x x x y 1 = e 2 , y 10 = 21 e 2 and y 100 = 14 e 2 . x x x x 1 x 1 x 4y 100 − 4y 10 + y 1 = 4 · e 2 − 4 · e 2 + e 2 = e 2 − 2e 2 + e 2 = 0. 4 2 • Check that y 2 is a solution of 4y 00 − 4y 0 + y = 0. x x x x x x x x y 2 = xe 2 , y 20 = e 2 + 21 xe 2 and y 200 = 12 e 2 + 21 e 2 + 14 xe 2 = e 2 + 41 xe 2 . x x x x x x x x 1 x 1 x 4y 200 − 4y 20 + y 2 = 4(e 2 + xe 2 ) − 4(e 2 + xe 2 ) + xe 2 = 4e 2 + xe 2 − 4e 2 − 2xe 2 + xe 2 = 0. 4 2 • Check y 1 and y 2 are linearly independent. ¯ ¯ y1 w(y 1 , y 2 ) = ¯¯ 0 y1 ¯ ¯ x y 2 ¯¯ ¯¯ e 2 = x y 20 ¯ ¯ 12 e 2 ¯ x ¯ x x x x x 1 x x x xe 2 e 2 ) − xe 2 e 2 = e x + e x − e x = e x 6= 0 x x ¯ = e 2 (e 2 + x ¯ e 2 + 2e 2 2 2 2 2 on (−∞, ∞). Therefore, y 1 and y 2 are linearly independent. Hence y 1 and y 2 form a fundamental set of solutions for the above DE on (−∞, ∞). (b) [2 points]using y 1 and y 2 in part (a), form the general solution of the above differential equation. x x Solution: y = c 1 y 1 + c 2 y 2 = c 1 e 2 + c 2 xe 2 . 3 Question 3. [21 points] Find the general solution of the given second-order differential equation. (a) [7 points] y 00 − 3y 0 + 2y = 0 Solution: From m 2 −3m+2 = 0 we have (m−2)(m−1). So m 1 = 1 and m 2 = 2. So y = c 1 e x +c 2 e 2x . (b) [7 points] y 00 − 10y 0 + 25y = 0 Solution: From m 2 − 10m + 25 = 0 we have (m − 5)2 = 0. So m 1 = m 2 = 5 and y = c 1 e 5x + c 2 xe 5x . (c) [7 points] 2y 00 + 2y 0 + y = 0 Solution: From 2m 2 + 2m + 1 = 0 we have that p p 1 1 −2 ± 22 − 4 · 2 · 1 −2 ± 4 − 8 −2 ± 2i = = = − ± i. m1 , m2 = 2·2 4 4 2 2 ¢ x¡ So α = − 12 and β = 12 . Therefore, y = e − 2 c 1 cos 12 x + c 2 sin 12 x . 4 Question 4. [10 points] Find the general solution of the given third-order differential equation d 3x d 2x − − 4x = 0 dt3 dt2 Solution: From m 3 − m 2 − 4 = 0 we have that since a 0 = −4 it has factors of ±4, ±2, ±1 and since a 3 = 1 it has factors of ±1. p Check m = q , where p is the factor of a 0 and q is the factor of a 3 . We find that m = 2 is one root. Now, using long division, we know that m 3 −m 2 −4 = 0 ⇔ (m −2)(m 2 +m +2) = 0. Therefore, m 1 = 2 or m 2 + m + 2 = 0 in which case p p −1 ± 1 − 4 · 2 −1 ± 7i = m2 , m3 = 2 2 and α = − 21 and β = p 7 2 . 2t Therefore, x = c 1 e t + e − 2 (c 2 cos p 7 2 t + c 3 sin p 7 2 t ). 5 Question 5. [25 points] Solve the given differential equation by undetermined coefficients y 00 − 4y = (x 2 − 3) sin 2x Solution: For y 00 −4y = 0, we have m 2 −4 ⇒ m 2 = 4 ⇒ m 1 = 2, m 2 = −2. Then y c = c 1 e 2x +c 2 e −2x . Since g (x) = (x 2 − e) sin 2x we assume that y p = (Ax 2 + B x +C ) cos 2x + (E x 2 + F x +G) sin 2x. Now, y p0 = (2Ax + B ) cos 2x + (Ax 2 + B x +C )(−2 sin 2x) + (2E x + F ) sin 2x + (E x 2 + F x +G)(2 cos 2x) = (2Ax + B ) cos 2x + (−2)(Ax 2 + B x +C ) sin 2x + (2E x + F ) sin 2x + 2(E x 2 + F x +G) cos 2x. y p00 = 2A cos 2x + (2Ax + B )(−2 sin 2x) + (−2)(2Ax + B ) sin 2x + (−2)(Ax 2 + B x +C )(2 cos 2x) + 2E sin 2x + (2E x + F )(2 cos 2x) + 2(2E x + F ) cos 2x + 2(E x 2 + F x +G)(−2 sin 2x). From y p00 − 4y p = (x 2 − 3) sin 2x we have: For x 2 cos 2x: −4A − 4A = −8A = 0 ⇒ A = 0. For x cos 2x: −4B + 4E + 4E − 4B = −8B + 8E = 0 ⇒ B = E = − 18 . For cos 2x: 2A − 4C + 2F + 2F − 4C = 2A − 8C + 4F = 0 ⇒ C = 0. For x 2 sin 2x: −4E − 4E = −8E = 1 ⇒ E = − 18 . For x sin 2x: −4A − 4A − 4F − 4F = −8A − 8F = 0 ⇒ F = 0. 13 For sin 2x: −2B − 2B + 2E − 4G − 4G = −4B + 2E − 8G = −3 ⇒ G = 32 . ) sin 2x. Therefore, the general solution is y = c 1 e 2x +c 2 e −2x − Hence y p = − 18 x cos 2x +(− 81 x 2 + 13 32 1 1 2 13 x cos 2x + (− 8 x + 32 ) sin 2x. 8 6 Question 6. [20 points] Solve the given initial-value problem 5y 00 + y 0 = −6x, y(0) = 0, y 0 (0) = −10 Solution: For 5y 00 + y 0 = 0 we have 5m 2 + m = 0 ⇒ m(5m + 1) = 0 ⇒ m 1 = 0, m 2 = − 15 . x So y c = c 1 + c 2 e − 5 . Since g (x) = −6x we first assume y p = Ax + B . However, B here is duplicated in y c . Then the correct assumption for y p is y p = Ax 2 + B x. So, y p0 − 2Ax + B and y p00 = 2A. So 5y p00 + y p0 = 5 · 2A + 2Ax + B = 2Ax + (10A + B ). We have, 2Ax + (10A + B ) = −6x. Therefore 2A = −6 ⇒ A = −3 and 10A + B = 0 ⇒ B = 30. Hence, y p = −3x 2 + 30x. x The general solution is y = c 1 + c 2 e − 5 − 3x 2 + 30x. Now applying I.C. to y and y 0 gives y 0 = x − 51 c 2 e − 5 − 6x + 30. y(0) = 0 ⇒ 0 = c 1 + c 2 e 0 ⇒ c 1 + c 2 = 0 ⇒ c 1 = −200. y 0 (0) = −10 ⇒ −10 = − 51 c 2 e 0 − 0 + 30 ⇒ c 2 = 200. x Therefore, the solution of the initial value problem is y = −200 + 200e − 5 − 3x 2 + 30x. 7 Question 7. [22 points] Solve the given differential equation by variation of parameters 3y 00 − 6y 0 + 6y = e x sec x Solution: 1. For 3y 00 − 6y 0 + 6y = 0 we have 3m 2 − 6m + 6 = 0 ⇒ 3(m 2 − 2m + 2) = 0. So p 2 ± 4 − 4 · 2 2 ± 2i m1 , m2 = = = 1±i 2 2 and α = 1 = β. So, y c = e x (c 1 cosx + c 2 sin x). Hence y 1 = e x cos x and y 2 = e x sin x. 2. We have ¯ ¯ y1 w(y 1 , y 2 ) = ¯¯ 0 y1 ¯ ¯ y 2 ¯¯ ¯¯ e x cos x e x sin x = y 20 ¯ ¯ e x cos x − e x sin x e x sin x + e x cos x ¯ ¯ ¯ ¯ = e x cos x(e x sin x + e x cos x) − e x sin x(e x cos x − e x sin x) = e 2x cos x sin x + e 2x cos2 x − e 2x sin x cos x + e 2x sin2 x = e 2x (cos2 x + sin2 x) = e 2x . 3. 3y 00 − 6y 0 + 6y = e x sec x ⇒ y 00 − 2y 0 + 2y = 13 e x sec x. f (x) = 31 e x sec x. 4. We have ¯ ¯ 0 w 1 = ¯¯ f (x) ¯ ¯ y1 w 2 = ¯¯ 0 y1 ¯ ¯ ¯ ¯ 0 e x sin x y 2 ¯¯ ¯¯ ¯ = − 1 e 2x sin x sec x 0 ¯=¯ 1 x x x ¯ y2 e sec x e sin x + e cos x 3 3 ¯ ¯ ¯ x ¯ ¯ ¯ e cos x 0 0 ¯ ¯ ¯ = 1 e 2x cos x sec x. =¯ x 1 x x ¯ f (x) e cos x − e sin x e sec x ¯ 3 3 So 1 2x w 1 − 3 e sin x sec x 1 1 1 1 = = − tan x. = − sin x sec x = − sin x 2x w e 3 3 cos x 3 1 2x w 2 3 e cos x sec x 1 1 u 20 = = = cos x sec x = . 2x w e 3 3 u 10 = Therefore, Z Z Z Z 1 1 1 sin x 1 sin x 1 1 u 1 = − tan xd x = − dx = − dx = d (cos x) = ln | cos x|. 3 cos x 3 cos x 3 cos x 3 Z 3 1 1 d x = x. u2 = 3 3 5. y p = u 1 y 1 + u 2 y 2 = 31 e x cos x ln | cos x| + 31 xe x sin x. 8 Question 8. [BONUS: 21 points] Discuss how the methods of undetermined coefficients and variation of parameters can be combined to solve the given differential equation. Carry out your idea. y 00 − 2y 0 + y = 4x 2 − 3 + x −1 e x [HINT: Superposition Principle – Nonhomogeneous Equations] Solution: For y 00 − 2y 0 + y = 0 the auxiliary equation is m 2 − 2m + 1 − 0 ⇒ (m − 1)2 = 0 ⇒ m 1 = m 2 = 1. So, y c = c 1 e x + c 2 xe x . Now, we consider first the DE y 00 −2y 0 + y = 4x 2 −3 which can be solved using undetermined coefficients where g 1 (x) = 4x 2 − 3. Then y p 1 = Ax 2 + B x + C and then substituting into the DE we get Ax 2 + (−4A + B )x + (2A − 2B +C ) = 4x 2 − 3. Then A = 4, −4A + B = 0 and 2A − 2B + C = −3. SO A = 4, B = 16 AND C = 21. Thus y p 1 = 4x 2 + 16x + 21. Next, we consider the DE y 00 − 2y 0 + y = x −1 e x for which a particular solution y p 2 can be x found using variation of parameters. The Wronskian is W = e 2x . Identifying f (x) = ex we obtain u 10 = −1 and u 20 = x1 . Then u 1 = −x and u 2 = ln x so that y p 2 = −xe x + xe x ln x. The general solution of the original DE is then y = yc + y p1 + y p2 = c 1 e x + c 2 xe x + 4x 2 + 16x + 21 − xe x + xe x ln x = c 1 e x + c 3 xe x + 4x 2 + 16x + 21 + xe x ln x. 9