Re: [理工] [工數]-高階ODE

[suker (..)]

※ 引述《roud (對愛絕望)》之銘言：
:      y'' + sin y = 0 , y(0) = 0, y'(0) = 2
: ans: ln | sec y/2 + tan y/2 | = x
: 拜託了~感恩

P=y'
            d P          d P        d y         d P
P' =y'' = -------  =  --------- *---------  =P ------
            d x          d y        d x         d y


   d P
P -------  +   sin y = 0
   d y


∫ P d P  =  ∫ -siny dy


P^2  /2   =  Cos y  + C


(y')^2  =  2  Cos y  +C1

y'(0) = 2

   ==>C1=2

(y')^2  =  2  (Cos y+ 1)   =  4*(cos(y/2))^2



  (y')   =   2 cos(y/2)     or     (y')   =   -2 cos(y/2)
      y'(0) = 2                  (y')   =   -2 cos(y/2)  不合



======================>
     (y')   =   2 cos(y/2)


     dy /dx   =   2 cos(y/2)

        dx =  (1/2) *sec(y/2)  dy

       x=      ln | sec y/2 + tan y/2 |  +C2

   y(0) = 0  ==> C2 =0


           x=      ln | sec y/2 + tan y/2 |


























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◆ From: 118.169.80.127
※ 編輯: suker           來自: 118.169.80.127       (07/09 19:42)
推 roud:謝謝~