MAT 2384 3X DIFFERENTIAL EQUATIONS AND NUMERICAL METHODS Final Exam July 25, 2012 Instructor: Dr. Steve Desjardins Duration: 3 hours Name: Student Number: Instructions: - Print your name and student number on this page. - Verify that your copy of the exam has all 11 pages. - You must answer all questions. There are 8 questions worth a total of 60 marks. - Write your answers in the spaces below the questions. You may use the backs of the pages if necessary. Use the backs of the formula pages for rough work if needed. - There are 2 pages of formulas (which may be detached) at the end of the exam. - No Notes or Books. - Basic scientic calculators only - graphing and/or programmable calculators are NOT permitted. MAT 2384 3X S12 - Final Exam 2 Question 1 (7 marks) Solve the initial value problem: (y 2 + y sin x cos y) dx + (xy + y cos x sin y) dy = 0, y(0) = /2 . MAT 2384 3X S12 - Final Exam 3 Question 2 (8 marks) Solve the initial value problem: y 9y = 54x 9 20e2x , y(0) = 8, y (0) = 5, y (0) = 38 . MAT 2384 3X S12 - Final Exam 4 Question 3 (8 marks) Solve the initial value problem: x2 y 4xy + 6y = x3 , x > 0, y(1) = 3, y (1) = 9 . MAT 2384 3X S12 - Final Exam Question 4 (8 marks) Find the general solution of the nonhomogeneous system: y1 = y2 + 4 y2 = 9y1 + 6y2 + 9x 5 MAT 2384 3X S12 - Final Exam Question 5 (8 marks) Find the following: (a) L{tet sin(2t)} (b) L{e3t sin(2t)} (c) L{u(t 1) (t2 + 3t)} (d) L1 2s + 5 s2 + 2s + 10 6 MAT 2384 3X S12 - Final Exam 7 Question 6 (7 marks) Use the Laplace Transform to solve the initial value problem: y + 4y + 3y = (t 3), y(0) = 4, y (0) = 6 . MAT 2384 3X S12 - Final Exam 8 Question 7 (7 marks) 1 2 dx to 6 decimal places. 2 1 1 + x Compare the approximation with the true value by calculating the simple error, ie |true - approx| . Use Gaussian Quadrature with 4 steps to approximate MAT 2384 3X S12 - Final Exam 9 Question 8 (7 marks) Use the Runge-Kutta Method of order 4 with h = 0.2 to calculate (to 4 decimal places) the rst two steps (ie y1 and y2 ) of the numerical solution of y = 2xy , y(0) = 4 . Compare the approximations with the true values by calculating the simple errors, ie |true - approx| . MAT 2384 3X S12 - Final Exam 10 Formulas f (t) F (s) = L{f (t)} n t n!/sn+1 at 1/(s a) e sin(kt) k/(s2 + k 2 ) s/(s2 + k 2 ) cos(kt) k/(s2 k 2 ) sinh(kt) cosh(kt) s/(s2 k 2 ) eas (t a) eas u(t a) s ; ; ; ; ; ; ; ; n = 0, 1, 2, . . . and s > 0 s>a s>0 s>0 s>k s>k s>0 s>0 est f (t)dt L{f (t)}(s) = 0 at L{e f (t)} = F (s a) L{u(t a)f (t a)} = eas F (s) dn L{tn f (t)} = (1)n n F (s) ds t 1 f (x)dx = F (s) L s 0 f (t) L F (x)dx = t s dn L f (t) = sn F (s) sn1 f (0) sn2 f (0) . . . f (n1) (0) dtn t f (x)g(t x)dx (f g)(t) = 0 L{(f g)(t)} = F (s) G(s) n b f (x ) , j f (x)dx = h a b f (x)dx = a b f (x)dx = a h 2 h 3 j=1 n | | 1 M (b a) h2 , 24 (f (xj1 ) + f (xj )) , j=1 | | M = maxaxb |f (x)| 1 M (b a) h2 , 12 n1 (f (x2j ) + 4f (x2j+1 ) + f (x2j+2 )) , j=0 M = maxaxb |f (4) (x)| | | M = maxaxb |f (x)| 1 M (b a) h4 , 180 MAT 2384 3X S12 - Final Exam 11 P C C yn+1 = yn + h f (xn , yn ) 1 C C C P yn+1 = yn + h f (xn , yn ) + f (xn+1 , yn+1 ) 2 k1 = h f (xn , yn ) 1 1 k2 = h f (xn + h, yn + k1 ) 2 2 1 1 k3 = h f (xn + h, yn + k2 ) 2 2 k4 = h f (xn + h, yn + k3 ) 1 yn+1 = yn + (k1 + 2 k2 + 2 k3 + k4 ) 6 Order n 2 3 4 5 Nodes ti -0.5773502692 0.5773502692 -0.7745966692 0.0 0.7745966692 -0.8611363116 -0.3399810436 0.3399810436 0.8611363116 -0.9061798459 -0.5384693101 0.0 0.5384693101 0.9061798459 Coecients Ai 1.0 1.0 0.555555556 0.888888889 0.555555556 0.3478548451 0.6521451549 0.6521451549 0.3478548451 0.2369268850 0.4786286705 0.5688888889 0.4786286705 0.2369268850