1. 耍人呀 = = 係數應該沒y吧... 2 3 x y" - x(x+2)y' + (x+2)y = 2x 觀察得一齊性解為x Let y = xu , y' = u + xu' , y" = 2u' + xu" 2 3 x (2u'+xu") - x(x+2)(u+xu') + (x+2)xu = 2x 除以x並且乘開 2 2 2 2xu' + x u" - xu - 2u - x u' - 2xu' + xu + 2u = 2x 2 2 2 x u" - x u' = 2x Let p = u' p' - p = 2 dp ---- = 2 + p dx dp ----- = dx 2+p ln|p+2| = x + c* x p+2 = c1 e x p = u' = c1 e - 2 x du = (c1 e - 2) dx x u = c1 e - 2x + c2 x 2 y = xu = c1 xe + c2 x - 2x 2. t 2e y" - y = --------- t -t e + e 先求齊性解 mt Let y=e 2 m - 1 = 0 m = 1 or -1 t -t yh = c1e + c2e t t 2 e 1 1 e yp = --------- --------- = (------- - -------) --------- 2 t -t D - 1 D + 1 t -t D - 1 e + e e + e 2t t 1 1 -t 1 e = e --- --------- - e --- --------- D t -t D t -t e + e e + e t t -t -t t e -t ( e + e ) - e = e ∫--------dt - e ∫-----------------dt 2t -2t e + 1 1 + e t t -t -t Let x = e , dx = e dt , z = e , dz = -e dt t 1 -t t 1 = e ∫-------dx - e ( ∫e dt - ∫------- dz ) 2 2 x + 1 z + 1 t -1 t -t -1 -t = e tan e - 1 + e tan e t -t t -1 t -t -1 -t y = yh + yp = c1e + c2e - 1 + e tan e + e tan e 3. ∫(xlnx - x)dx 2 1 Let u = x , du = 2xdx , v = lnx , dv = ---dx x 1 1 = ---∫(v-1)du = ---[ u(v-1) - ∫udv ] 2 2 1 2 1 2 3 2 = ---[ x (lnx - 1) - ∫xdx ] = --- x lnx - --- x 2 2 4 4. ix 1 Im[e ------------------ x] 2 D + 2(i+1)D + 2i 分母先提個2i 讓D的零次方項為1 ix e 1 = Im[----- ------------------------- x] 2i i+1 1 2 1 + (----- D + ---- D ) i 2i 1 2 代入泰勒展開式 ------- = 1 - t + t - … 1 + t 由於只作用在x的一次方上 取到D的一次方就夠了 ix e i+1 = Im[----- (1 - -----D)x] 2i i 2 所以剛剛分母那也可以直接把D 項去掉 答案亦同 5. y" + 4y = cos2x mx Let y = e 2 m + 4 = 0 , m = ±2i yh = c1 cos2x + c2 sin2x 1 yp = ------- cos 2x 2 D + 4 1 2ix = Re [------- e ] 2 D + 4 1 1 1 2ix = Re [---- (------ - ------)e ] 4i D-2i D+2i 1 2ix -2ix 4ix = --- Re [-i(xe - e ∫e dx )] 4 4ix 1 2ix -2ix e = --- Re [-i(xe - e -------)] 4 4i 1 1 = --- Re [-i x(cos2x + i sin2x) + ---(cos2x + i sin2x)] 4 4 1 1 = --- (xsin2x + ---cos2x) 4 4 1 y = yh + yp = c1 cos2x + c2 sin2x + ---xsin2x 4 --

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