Re: [理工] [工數]-高階ODE

[youmehim (哩挖伊)]

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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