Re: [理工] [工數]-變係數ODE

[youmehim (哩挖伊)]

3.
       2
y" = y' + y'y   ,   y(0) = 1   ,   y'(0) = -2

                    d   dy     dy    d   dy       dp
Let p = y' , y" = ---- ---- = ---- ---- ---- = p ----
                   dx   dx     dx   dy   dx       dy
       2
pp' = p + py (從這邊開始 '表示對y微分)

p(p' - p - y) = 0

(i)

p = 0 , y = c1 = 1

但不符初始條件 y'(0) = -2 (不合)

(ii)

p' - p = y

               -y
I= exp∫-dy = e

         -y        -y   -y
Ip = ∫ye  dy = -ye  - e  + c2

          1     -y   -y                     y
p = y' = ---(-ye  - e  + c2) = -y - 1 + c2 e
          I

將初始條件 y(0) = 1   ,   y'(0) = -2 代入

-2 = -1 - 1 + c2

得 c2 = 0

y' = -y-1

   dy
------- = -dx
 y + 1

ln|y+1| = -x + c3*

        -x
y = c3 e  - 1

將初始條件 y(0) = 1 代入

1 = c3 - 1

得c3 = 2

           -x
解為 y = 2e  - 1

5.
         2
yy" = 2y' - 2y' , y(0) = 1 , y'(0) = 2

                    dp
Let p = y' , y" = p----
                    dy

         2
ypp' = 2p - 2p (從這邊開始 '表示對y微分)



p(yp' - 2p + 2) = 0

(i)

p = 0 , y = c1 = 1

但不符合初始條件 y'(0) = 2 (不合)

(ii)

yp' - 2p + 2 = 0

      2         2
p' - --- p = - ---
      y         y

            2        -2
I = exp∫- --- dy = y
            y

          2   -2          -3      -2
Ip = ∫- --- y  dy = -2∫y  dy = y  + c2
          y
                 2
p = y' = 1 + c2 y

將初始條件 y(0) = 1   ,   y'(0) = 2 代入


2 = 1 + c2

得 c2 = 1

          2
y' = 1 + y

   dy
-------- = dx
      2
 1 + y

  -1
tan y = x + c3

將初始條件 y(0) = 1 代入

    π
得 --- = c3
    4
                  π
解為 y = tan(x + ---)
                  4

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※ 編輯: youmehim        來自: 218.173.129.116      (09/13 02:24)