Re: [其他] 請教一題工數題

[whalelover (哞哞哞哞王)]

Prob: y'' + 2y' + 2y = exp(-x)/(cosx)^3

Sol:

  解特徵方程式 m^2 + 2m + 2 = 0  =>  m = -1+i , -1-i

  整理後得 y_1 = exp(-x)cosx , y_2 = exp(-x)sinx

  W = | y_1  y_2 | = exp(-2x)
      | y_1' y_2'|

  By variation of parameters:

     y'' + P(x)y' + Q(x)y = f(x)之特解有以下形式

                                             y_2*f(x)          y_1*f(x)
     y_p = u_1*y_1 + u_2*y_2, where u_1' = - -------- , u_2' = --------
                                                W                 W

  所以 u_1' = -sinx/(cosx)^3 ,

       u_2' = 1/(cosx)^2 = (secx)^2

       u_1 = -1/2*(sec)^2 , u_2 = tanx

  => y_p = [-1/2*(sec)^2]*exp(-x)cosx + tanx*exp(-x)sinx

                             1              sinx
         = exp(-x)[-1/2 * -------- * cosx + ---- * sinx]
                          (cosx)^2          cosx

         = exp(-x)[-1/2*secx + secx*(sinx)^2]

         = -1/2*exp(-x)secx*cos(2x)
                                    ##

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感覺答案少寫後面那一塊吧?


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推 hahaiduno   : 請問cos(2x)怎麼整理得到的？不是-e^-x*cosx嗎？
cos(2x) = 1 - 2*(sinx)^2  只是我手癢再化簡而已 不要在意他

※ 編輯: whalelover (140.112.218.174), 01/07/2018 21:13:22