→ RedQ:排版用心,但是這樣根本沒解到，殘留那個積分式才是問題 06/24 11:46
→ RedQ:第一題相當於第二題，然而第二題你只是將他開根號加上積分 06/24 11:48
→ Frobenius:Mathematica跟我解的結果一樣(表示無法積分成初等函數) 06/24 12:00
※ 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
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◆ From: 218.166.106.225
※ 編輯: Frobenius 來自: 218.166.106.225 (06/24 03:58)