tut11_06_exam_questions

tut11_06_exam_questions - y (0) = 0 A = 0 y = Be-x sin( -1...

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1 MATH 3705 Final Examination Winter 2006 (Questions solved in last tutorial) 1. F{ e - 3 ix -| x - 2 | } = (b) (a) 2 e 2 i ( λ +3) 1 + ( λ + 3) 2 (b) 2 e 2 i ( λ - 3) 1 + ( λ - 3) 2 (c) 2 e - 2 i ( λ +3) 1 + ( λ + 3) 2 (d) 2 e - 2 i ( λ - 3) 1 + ( λ - 3) 2 (e) None of the above. Solution: F{ e - 3 ix -| x - 2 | } = F{ e - 3 ix · e -| x - 2 | } = { First Shift. Thm } = b f ( λ - 3), where b f ( λ ) = F{ e -| x - 2 | } = { Second Shift. Thm } = e 2 F{ e -| x | } = e 2 2 1 + λ 2 . Thus, b f ( λ - 3) = 2 e 2 i ( λ - 3) 1 + ( λ - 3) 2 . 2. F - 1 { e -| λ | } = (a) (a) 1 π 1 1 + x 2 (b) e -| x | (c) 2 1 + x 2 (d) 1 π e - x (e) None of the above. Just know the answer for this one – appears often on finals. It is derived from first principles (i.e. from the definition of F . Transform) and is not worth deriving each time.
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2 3. Find all eigenvalues and corresponding eigenfunctions of the Sturm-Liouville problem y 00 + 2 y 0 + λy = 0 , y (0) = 0 , y (1) = 0 Solution: y = e αx α 2 + 2 α + λ = 0 α = - 2 ± 4 - 4 λ 2 = - 1 ± 1 - λ. λ > 1 : α = - 1 ± i λ - 1 y = e - x [ A cos( λ - 1 x ) + B sin( λ - 1 x )].
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Unformatted text preview: y (0) = 0 A = 0 y = Be-x sin( -1 x ). y (1) = 0 Be-1 sin( -1) = 0 -1 = n . Thus, n = n 2 2 + 1 , n 1, are the eigenvalues, and y n = B n e-x sin( nx ) are the corresponding eigenfunctions. = 1 : =-1 y = e-x [ A + Bx ]. y (0) = 0 A = 0 y = Bxe-x . y (1) = 0 Be-1 = 0 B = 0 y = 0. < 1 : =-1 1- y = Ae (-1+ 1- ) x + Be (-1- 1- ) x . y (0) = y (1) = 0 A + B = A e (-1+ 1- ) + B e (-1+ 1- ) = A = B = 0, since 1 1 e (-1+ 1- ) e (-1+ 1- ) 6 = 0....
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This note was uploaded on 07/16/2009 for the course MATH 3705 taught by Professor Jaberabdualrahman during the Winter '08 term at Carleton CA.

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tut11_06_exam_questions - y (0) = 0 A = 0 y = Be-x sin( -1...

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