# 285hw5 - HW 5 1 Sec 3.3:11.1 y(4 8y(3 16y 00 = 0...

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HW 5 1. Sec. 3.3:11. 1 y (4) 8 y (3) + 16 y 00 = 0 : Characteristic equation: r 4 8 r 3 + 16 r 2 = 0 ) r 2 r 2 8 r + 16 ± = 0 ) r 1 = r 2 = 0; r 3 = r 4 = 4 : General solution: y ( x ) = ( C 0 + C 1 x ) + ( C 3 + C 4 x ) e 4 x : 2. Sec. 3.3: 25. 3 y (3) + 2 y 00 = 0 ; y (0) = 1 ; y 0 (0) = 0 ; y 00 (0) = 1 : Characteristic equation: 3 r 3 + 2 r 2 = 0 ) r 2 (3 r + 2) = 0 ) r 1 = r 2 = 0 ; r 3 = 2 3 : General solution: y ( x ) = C 1 + C 2 x + C 3 e 2 3 x : Now we use the initial value conditions: C 1 + C 3 = 1 C 2 2 3 C 3 = 0 4 9 C 3 = 1 Solution C 1 = 13 4 ; C 2 = 3 2 ; C 3 = 9 4 . Answer: y ( x ) = 13 4 + 3 2 x + 9 4 e 2 3 x : 1 Note to the grader. Please grade all 5 problems: Sec. 3.3:11, Sec. 3.3: 25, Sec. 3.3: 27, Sec. 3.3: 31, Sec. 3.3: 49 (20 pts for each problem). 1

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3. Sec. 3.3: 27. y (3) + 3 y 00 4 y = 0 : Characteristic equation: r 3 + 3 r 2 4 = 0 : Try r = ± 1 ; ± 2 ; ± 4 . Let r = 1 . Then 1 + 3 4 = 0 . Hence r = 1 is a root. r
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285hw5 - HW 5 1 Sec 3.3:11.1 y(4 8y(3 16y 00 = 0...

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