作者youmehim (哩挖伊)
看板Grad-ProbAsk
標題Re: [理工] [工數]-變係數ODE
時間Fri Sep 4 00:19:58 2009
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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◆ From: 58.114.111.147
1F：推 mdpming:...@@太利害了 09/04 00:24
2F：推 CRAZYAWIND:囧 我第三題只算到P = 後面那一串卡在沒想到要先代初始 09/04 00:28
3F：推 kusorz:原來要先帶條件阿 我想說最後不知道怎麼積 09/04 00:28
4F：推 CRAZYAWIND:多學到一招可以用的了= = 09/04 00:30
5F：推 hihaka2001:哈真的學到新招 09/04 10:19
※ 編輯: youmehim 來自: 218.173.129.116 (09/13 02:24)