※ 引述《roud (對愛絕望)》之銘言：
: y y'' = 2(y')^2 - 2 y' , y(0) = 1, y'(0) = 2
: ans: y = tan(x + π/4)
: 拜託了~感激
大同小異吧 P=y' ,P' =y'' =P dP/dy
P y dP/dy =2P^2 -2P (奇異解 P=0 =>y=1 )
y'=0 不符合 y'(0) = 2 不合
P 不等於0 約掉
ydP/dy =2(P-1)
1/ y dy = 1/2(P-1) dP
ln y = (1/2) ln (P-1) +C
2 ln y= ln (P-1) +2C
2ln y = ln (P-1) +ln C1
y^2 = C1*(P-1)
y^2 = C1* (y' -1)
y(0) = 1, y'(0) = 2
1=C1*(2-1) ==>C1=1
y^2 = y' -1
dy/dx = y^2 +1
dx = 1/ (y^2 +1) dy
x= arctan (y) + C2
y(0) = 1
0 = π/4 +C2 ===>C2 =-π/4
這邊會有1個會有 arctan (1)問題 tan k =1
k=π/4 +2nπ (n為整數)
或5π/4+2nπ (n為整數)
我習慣還是寫π/4
arctan (y) =x+π/4
===>y= arctan(x+π/4)
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◆ From: 118.169.78.115
※ 編輯: suker 來自: 118.169.78.115 (07/10 09:03)