Re: [工數] Nonlinear Equation一題

[yueayase (scrya)]

※ 引述《EZ0928 (傲熊)》之銘言：
: y''+(y')^2+1=0
: solve the given differential equation by using u=y'


Using chain rule:

       du     du   dy       du
y'' = ---- = ---- ---- = u ----
       dt     dy   dt       dy


      du     2
=> u ---- + u + 1 = 0
      dy

       2u
=>  -------- du = - 2 dy
      2
    (u + 1)

         2                 2
=>  ln(u + 1) = - 2y + ln C

    2    2 -2y
=> u  = C e   - 1

         dy      2 -2y   1/2
=> u  = ---- = (C e  - 1)
         dt

       y
      e
=> ------------- dy  =  dt
      2   2y 1/2
    (C - e  )

         y           y
Let v = e   => dv = e  dy


          dv
=>  ------------ = dt
       2   2 1/2
     (C - v )


       -1 v
=>  sin  --- = t + D
          C

     y
v = e

      y
So,  e = C sin(t+D)


當然可能前面要加正負號(剛才去平方根只有取正的)

其實形如 f(y'',y',y) = 0 的微分方程都是固定這樣做(令u=y')

關鍵是用chain rule去把y當成微方的independent variable

                     du
=> f(y'',y',y) = f(u----,u,y) = 0 變為一階
                     dy


f(y'',y',t) = 0 也是 令u=y'

只是這個比較簡單

直接將u對t微分就可

=>f(y'',y',t) = f(u',u,t) = 0 變為一階

這兩個都是 Reducible Second-Order Differential Equation

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→ yueayase    :啊,想得太複雜了
推 Frobenius   :推 這樣解也可以
※ 編輯: yueayase        來自: 111.251.163.20       (12/05 02:26)
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