# 285hw8 - HW 81 1 Sec 3.8:1 We have y 00 y = 0 y 0(0 = 0 y(1...

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HW 8 1 1. Sec. 3.8:1. We have y 00 + &y = 0 ; y 0 (0) = 0 ; y (1) = 0 : If & = 0 then we have y 00 = 0 = ) y 0 = a = ) y ( x ) = ax + b: Now we use the endpoint conditions: y 0 (0) = a = 0 = ) a = 0 ; y (1) = a + b = 0 = ) b = 0 : Thus, if & = 0 = ) y ( x ) = 0 = ) & = 0 is not eigenvalue. If & > 0 = ) & = ± 2 > 0 ; so we have y 00 + ± 2 y = 0 : The characteristic equation is r 2 + ± 2 = 0 = ) r = & ±i: The solution has the following form y ( x ) = A cos ±x + B sin ±x; then y 0 ( x ) = ± sin ±x + cos ±x: Now we use the endpoint condi- tions: y 0 (0) = = 0 = ) B = 0; y (1) = A cos ± + B sin ± = 0 = ) A cos ± = 0 = ) cos ± = 0 : Thus ± = & 2 + ²n = & 2 (1 + 2 n ) ; n ² 0; then & n = ± 2 = ² 2 4 (1 + 2 n ) 2 ; n ² 0; and the associated functions are y n ( x ) = cos ²x 2 (1 + 2 n ) ; n ² 0 : 2. Sec. 3.8:3. We have y 00 + &y = 0 ; y ( ± ² ) = 0 ; y ( ² ) = 0 : If & = 0 then we y 00 = 0 = ) y 0 = a = ) y ( x ) = ax + b . Now we use the endpoint conditions: y ( ± ² ) = ± + b = 0 ; y ( ² ) = + b = 0 : 1 Note to the grader. Please grade the following problems: Sec. 3.8:1, Sec. 3.8:5, Sec. 3.8:8, Sec. 9.1: 3, Sec. 9.1:13 (20 pts for each problem) 1

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So, 2 b = 0 = ) b = 0 = ) a = 0 : Thus, if & = 0 = ) y ( x ) = 0 = ) & = 0 is not eigenvalue. If & > 0 = ) & = ± 2 > 0 ; then we have y 00 + ± 2 y = 0 : The characteristic equation is r 2 + ± 2 = 0 = ) r = & ±i: The solution has the following form y ( x ) = A cos ±x + B sin ±x: Now we use the endpoint conditions: y ( ± ² ) = A cos ±² ± B sin ±² = 0 ; y ( ² ) = A cos ±² + B sin ±²
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## This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285hw8 - HW 81 1 Sec 3.8:1 We have y 00 y = 0 y 0(0 = 0 y(1...

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