[复变函数与积分变换].焦红伟&尹景本.文字版.PDF.pdf

F t t δ 14 1 1 sin sin 2 3 6 f t t t 15 2 e e t t f

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f t t δ = (14) 1 1 ( ) sin sin 2 3 6 f t t t = (15) 2 e e ( ) t t f t t = (16) 1 ( ) e (sin cos ) 2 t f t t t t = (17) ( ) 5sin 2 cos2 f t t t = 7 (1) 3 1 ( ) [(7 2 )e 3e ] 4 t t y t t = + (2) 2 ( 1) 2( 1) 1 1 ( ) e e [ e e ] ( 1) 2 2 t t t t y t u t = + − + + (3) ( ) 2sin cos2 y t t t = − (4) ( ) e sin t y t t t = (5) 2 1 1 1 ( ) e sin 2 10 5 t y t t = − + (6) 1 ( ) sin 2 y t t t = 8 2 2 2 2 2(3 12 13) ( ) [( 3) 4] s s F s s s + + = + + 9 3 ( ) 6 t f t a t = + 10 .解:设 [ y ( t ) ]= Y ( s ) ,对方程两边同取拉氏变换.并考虑到初始条件,得到
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习题答案 · 163 · · 163 · 2 1 ( ) 1 2 ( ) 3 ( ) 1 s Y s sY s Y s s + = + 整理得 2 ( ) ( 1)( 1)( 3) s Y s s s s + = + + 根据海维赛展开式来求其逆变换.这里 B (s) 有三个单零点: 1 2 3 1, 1, 3 p p p = − = = − . 2 2 2 1 1 3 2 2 2 ( ) e e e 3 6 1 3 6 1 3 6 1 st st st s s s s s s y t s s s s s s =− = =− + + + = + + + + + 3 3 1 3 1 1 e e e (3e 2e e ) 4 8 8 8 t t t t t t = − + = 即为所求微分方程的解. 11 ( ) k x t t m =
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  • Winter '16
  • Gilles Lamothe

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