# 285hw4 - 6 Sec 3.2 13 y(3 2 y 00& y& 2 y = 0 y(0 = 1...

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HW 4 1 1. Sec. 3.1: 25. f ( x ) = e x sin x; g ( x ) = e x cos x . Wronskian: W [ f; g ] = e x sin x e x cos x e x sin x + e x cos x e x cos x e x sin x : Now calculate W [ f; g ] j x =0 = & 0 1 1 1 = 1 6 = 0 , hence f and g are linearly independent. 2. Sec. 3.1: 37. 2 y 00 y 0 y = 0 : Characteristic equation: 2 r 2 r 1 = 0 ) r 1 = 1 ; r 2 = 1 2 . General solution: y ( x ) = C 1 e x + C 2 e 1 2 x . 3. Sec. 3.2: 1. f ( x ) = 2 x; g ( x ) = 3 x 2 ; h ( x ) = 5 x 8 x 2 : Let 2 x + ± 3 x 2 + ² ± 5 x 8 x 2 ² = 0 ; whence (2 + 5 ² ) x + (3 ± 8 ² ) x 2 = 0 : Hence we have 2 + 5 ² = 0 3 ± 8 ² = 0 Take ² = 1 , then ± = 8 = 3 and = 5 = 2 . Then 2 x ( 5 = 2) + 3 x 2 (8 = 3) + ± 5 x 8 x 2 ² ± 0 : Therefore the functions f ( x ) = 2 x; g ( x ) = 3 x 2 ; h ( x ) = 5 x 8 x 2 are linearly dependent. 1 Note to the grader. Please grade the following problems: Sec. 3.1: 25, Sec. 3.1: 37, Sec. 3.2: 1, Sec. 3.2: 5, Sec. 3.2: 21 (20p pts for each problem). 1

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4. Sec. 3.2: 5. Let f ( x ) = 17 ; g ( x ) = cos 2 x; h ( x ) = cos 2 x . Because 2 cos 2 x = 1 cos 2 x , we can write 1 17 f ( x ) + ( 1) h ( x ) + ( 2) g ( x ) ± 0 : The functions f ( x ) ; g ( x ) and h ( x ) are linearly dependent. 5. Sec. 3.2: 9.We assume that e x ; cos x; sin x are solutions of a di/erential equation of the second order. W [ e x ; cos x; sin x ] j x =0 = e x cos x sin x e x sin x cos x e x cos x sin x j x =0 = 1 1 0 1 0 1 1 1 0 = 2 : Hence e x ; cos x; sin x are linearly independent.
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Unformatted text preview: 6. Sec. 3.2: 13. y (3) + 2 y 00 & y & 2 y = 0; y (0) = 1 ; y (0) = 2 ; y 00 (0) = 0 ; y 1 = e x ; y 2 = e & x ; y 3 = e & 2 x . General solution: y ( x ) = C 1 e x + C 2 e & x + C 3 e & 2 x . Then, by using the initial conditions, we have C 1 + C 2 + C 3 = 1 C 1 & C 2 & 2 C 3 = 2 C 1 + C 2 + 4 C 3 = 0 Solution is : ± C 2 = 0 ; C 3 = & 1 3 ; C 1 = 4 3 ² . Hence y ( x ) = 4 3 e x & 1 3 e & 2 x : 7. Sec. 3.2: 21. y 00 + y = 3 x ; y (0) = 2 ; y (0) = & 2 :y c ( x ) = c 1 cos x + c 2 sin x ; y p ( x ) = 3 x: General solution: y ( x ) = c 1 cos x + c 2 sin x + 3 x: 2 By using the initial conditions, c 1 = 2 c 2 + 3 = & 2 : Hence c 1 = 2 and c 2 = & 5 . Answer: y ( x ) = 2 cos x & 5 sin x + 3 x . 3...
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285hw4 - 6 Sec 3.2 13 y(3 2 y 00& y& 2 y = 0 y(0 = 1...

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