Re: [工數] 解聯立方程式

[ntust661 (常春藤 i pod)]

※ 引述《kilikolo0218 (小霸王)》之銘言：
: 解聯立方程式
:    z''+ y' = cosx    y'(0)=1,y(0)=1
: {
:    y''- z = sinx     z'(0)=z(0)=-1
: 感謝解答



   y' = cosx - z''


   y'' = -sinx - z'''  代入(2)


   -sinx - z''' - z = sinx


   z''' + z = - 2sinx


     3
=> (D + 1)z = -2 sinx


          2
=> (D-1)(D + D + 1)z = -2sinx


homogenenous solution


            x     x/2
=> z  = c1 e  +  e   { c2 cos(√3/2)x + c3 sin(√3/2)x  }
    h


Particular solution


             2                  2                     2
=> z  = ───── sinx   ( -(1)  is substituted for D    )
    p    D^3 + 1


             2
      = ───── sinx
         -D + 1


        2 (D + 1)
     = ───── sinx  = (D+1) sinx = cosx + sinx
        1 - D^2



   a  = (√3/2)


          x    x/2
  z = c1 e  + e    { c2 cosax + c3 sinax } + cosx + sinx


  z(0) = -1 , z'(0) = -1


  z(0) = c1 + c2 + 1  = -1


  z'(0) = c1 + c2/2 + a c3 + 1 = -1


  =>

  c2/2 - ac3  = 0 , c2 = 2 a c3


  c1 = -2 - c2 = -2 - 2 a c3


  代回(1)


  z'' + y' = cosx


      x        x/2                              x/2
  c1 e  + 1/4 e    { c2 cosax + c3 sinax } + a e    { - c2 sinax + c3 cosax }


      2  x/2
  +  a  e   { -c2 cosax -c3 sinax} - cosx - sinx + y' = cosx



  y'(0) = 1

                       2
  c1 + c2/4 + -a c3 - a  - 1 + y'(0) = 1


                      2
  c1 + c2/4 - a c3 - a  = 1

                                     2                               2
  -2 - 2a c3 + a c3 /2 - a c3 = 1 + a   ,  (-2a + a/2 - a) c3 = 3 + a


          2           2
  c3 = ──── (3 + a )
         -5a

  ---------------------


  不要笨笨積分喔!!


  z'' + y' = cosx 對 x 積分就會有 y 了


  z' + y = sinx + C , y = sinx - z' + C


  y(0) = 0 - z'(0) + C =  1 + C = 1 , C = 0



  y(x) = sinx - z'

                      x        x/2                             x/2
       = sinx - [ c1 e  + 1/2 e   { c2 cosax + c3 sinax } + a e    {-c2 sinax


         + c3 cosax } + cosx - sinx ]

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