Re: [理工] [工數]-高階ODE
※ 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)
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07/09 20:18, , 1F
07/09 20:18, 1F
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