s251final(sp03)

s251final(sp03) - F(0 = F L = 0 Q:13 n =(2 n-1 2 4 f n t...

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Answers to Spring 2003 Final exam. Q#:1 Answer D. Q#:2 Answer C. Q#:3 Answer C. Q#:4 Answer B. Q#:5 Answer B. Q#:6 Answer A. Q#:7 Answer A. Q#:8 Answer C. Q#:9 A Solution exists on (0 , ). B The solution is y ( t )= c 1 t + c 2 1 t . Q#:10 A The linearized matrix is ± y - 3 x 2 y 2 x +2 ² . B The critical points are P =(0 , 0) and Q =( - 1 , 3). C Point P is a saddle point, which is unstable. Point Q is a centre, which is stable. Q#:11 A B C Part (a) has a cosine series and part (b) has a sine series. D ˜ f ( - 2) = 0, ˜ f ( 1 2 )= 1 2 and ˜ f (3) = - 3 2 . Q#:12 Denoting u ( x, y )= F ( x ) G ( y ), A G 00 ( y )+2 G 0 ( y ) - λy 2 G ( y )=0and F 00 ( x ) - λF ( x ) = 0 are the two ordinary differential equations. B The boundary conditions become
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Unformatted text preview: F (0) = F ( L ) = 0. Q#:13 λ n = (2 n-1) 2 4 f n ( t ) = sin ³ (2 n-1) t 2 ´ n ∈ N Q#:14 A u t = 2 u xx u x (0 , t ) = u x (10 , t ) = 0 u ( x, 0) = 3 t cos( πx 5 )-5 cos( π 2 ). B u ( x, t ) = 3 + exp(-2 π 2 t 25 ) cos( πx 5 ). C The steady state emperature is 3. Q#:15 A The general solution is X ( t ) = c 1 ± 1-1 ² e-t + c 2 ³± 1-1 ² te-t + ± 1 ² e-t ´ . B The specific solution is X ( t ) =-± 1-1 ² e-t + 3 ³± 1-1 ² te-t + ± 1 ² e-t ´ ....
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This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Penn State.

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