Math 293 prelim 3 solution

# Math 293 prelim 3 solution - Math 293 Prelim#3 Solutions...

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Math 293 Prelim #3 Solutions Spring 2000 1 . (a) π 2 π 3 π 3 π π - - 2 π - 0 (b) π 2 π 3 π 3 π π - - 2 π - 0 (c) b n = 2 π R π 0 sin x cos nx dx = 1 π R π 0 ( sin (n + 1 )x - sin (n - 1 )x) dx = 1 π ( - cos (n + 1 )x n + 1 + cos (n - 1 )x n - 1 π 0 ) = 1 π ( - cos (n + 1 n + 1 + 1 n + 1 + cos (n - 1 n - 1 - 1 n - 1 ) . This is 0 for n odd and 1 π ( 2 n + 1 - 2 n - 1 ) = - 4 π(n 2 - 1 ) for n even. When n = 1 this calculation involves dividing by 0, so we compute b 1 directly: b 1 = 2 π R π 0 sin x cos x dx = 2 π sin 2 x 2 π 0 = 0. It is not necessary to calculate a 0 separately since the calculation for a general value of n works when n = 0, and gives a 0 = 4 π . The final answer is 2 π - 4 π X n = 1 1 4 n 2 - 1 cos 2 nx . 2 . Case 1: λ = α 2 > 0. Then y = c 1 sin αx + c 2 cos αx , y 0 = αc 1 cos αx - αc 2 sin αx . We have y 0 ( 0 ) = αc 1 = 0 so c 1 = 0. Then y(π) = c 2 cos απ = 0, which implies α = 2 n - 1 2 for n = 1 , 2 , · · · . Thus we have the eigenvalues λ = ( 2 n - 1 2 ) 2 , with corresponding solutions y = c 2 cos ( 2 n - 1 2 ) x . Case 2: λ = 0, so y = c 1 x + c 2 . Then y 0 ( 0 ) = c 1 = 0, and y(π) = c 2 = 0. Thus we have only the trivial solution y = 0, and λ = 0 is not an eigenvalue.

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