※ 引述《roud (對愛絕望)》之銘言：
: y y'' = 2(y')^2 - 2 y' , y(0) = 1, y'(0) = 2
let y'=p y''=dp/dx=(dp/dy)×(dy/dx)=p*dp/dy 帶入原式
=>yp(dp/dy)=2p^2-2p
=>(2/y)dy = (p-1)
積分=> ln(y^2)+C1=ln(p-1)
∴p =C1y^2 + 1
帶入條件,when x=0,y(0) = 1, y'(0) = 2
得C1=1
∴p =y^2 + 1
=> dy/dx = y^2 +1
=> dy/(y^2 +1) = dx
積分 => arctan(y)= x+C2
=> y=tan(x+C2)
帶入條件得C2=π/4 & 5π/4(?)
得y = tan(x + π/4)
: ans: y = tan(x + π/4)
: 拜託了~感激
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