Re: [理工] [工數]-二階非線性ODE & 一階高次ODE
※ 引述《honestonly (努力增胖的小R)》之銘言：
: (1)
: 2
: d y
: a ---- + bcosy - c = 0; a, b, c是常數
: 2
: dt
:
: 求 y(t) = ?
Let z = y' = dy/dt =>
d^2 y/dt^2 = y'' = (y')' = z' = dz/dt = (dz/dy)(dy/dt) = z(dz/dy)
a z(dz/dy) + (bcosy - c) = 0
a 2zdz + 2(bcosy - c)dy = 0
a z^2 + 2(bsiny - cy) = a C1
z^2 + (2/a)(bsiny - cy) = C1
z^2 = C1 + (2/a)(cy - bsiny) = (dy/dt)^2
dt/dy = [ C1 + (2/a)(cy - bsiny) ]^(-1/2)
dt = [ C1 + (2/a)(cy - bsiny) ]^(-1/2) dy
y(t)
( t + C2 ) = ∫ [ C1 + (2/a)(cτ - bsinτ) ]^(-1/2) dτ
1
2 y(t) 2
( t + C2 ) = { ∫ [ C1 + (2/a)(cτ - bsinτ) ]^(-1/2) dτ }
1
: (2)
: dy 2
: cy - bsiny = a(----) ; a, b, c是常數
: dt
: 求 y(t) = ?
a z^2 = ( cy - bsiny )
z^2 = (1/a)( cy - bsiny ) = (dy/dt)^2
(dt/dy)^2 = [ (1/a)( cy - bsiny ) ]^(-1)
dt/dy = [ (1/a)( cy - bsiny ) ]^(-1/2)
dt = [ (1/a)( cy - bsiny ) ]^(-1/2) dy
y(t)
t+ C2 = ∫ [ (1/a)( cy - bsiny ) ]^(-1/2) dτ
1
2 y(t) 2
( t + C2 ) = { ∫ [ (1/a)(cτ - bsinτ) ]^(-1/2) dτ }
1
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.166.106.225
※ 編輯: Frobenius 來自: 218.166.106.225 (06/24 03:58)
→
06/24 11:46, , 1F
06/24 11:46, 1F
→
06/24 11:48, , 2F
06/24 11:48, 2F
→
06/24 12:00, , 3F
06/24 12:00, 3F
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 2 之 2 篇)：