Re: [理工] [工數]問一題ode

[honestonly (感冒了~哭哭)]

※ 引述《fatfatyu (泡麵王)》之銘言：
:   y^2
: ------ + 2ye^t+(y+e^t)y'=0
:    2
: y(0)=1
: 請問這題該怎麼解?

都沒有人回 那我來賺P幣好了 科科
    2
   y       t              t
( --- + 2ye ) dt + ( y + e ) dy = 0
   2

       t     t        1    2        2
[ 2yd(e ) + e dy ] + ---[ y dt + d(y ) ] = 0
                      2
    2t
d(ye  )    1     2        2
------- + --- [ y dt + d(y ) ] = 0
    t      2
   e

    2t     1     2   t     t   2
d(ye  ) + --- [ y d(e ) + e d(y ) ] = 0
           2

    2t    1     2 t
d(ye ) + --- d(y e ) = 0
          2

  2t   1   2 t
ye  + --- y e = c
       2

y(0) = 1

     1             3
1 + --- = c , c = ---
     2             2

  2t   1   2 t   3
ye  + --- y e = ---
       2         2

--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 118.167.136.139
※ 編輯: honestonly      來自: 118.167.136.139      (03/11 19:16)
→ mdpming:被搶先了~"~
推 laboy10:GOOD