推 topee:感謝 04/07 20:57
※ 引述《topee (eason)》之銘言:
: y'-y = y^2
: dy-ydx=y^2dx
: 除y^2
: y^-2dy - y^-1dx = dx
: 令u= y^1-2 =y^-1
: du = (1-2)y^-2dy
: -du-udx = dx
: du+udx = -dx
: (D+1)u = -1
: u = c1e^-x
: h
: u = (1-D)(-1)
: p
: = -1
: 1/y = c1e^-x-1
: y=(c1e^-x-1)^-1
好像不用白努力?
y'-y = y^2
dy
─ = y^2 + y = y(y+1)
dx
1
─── dy = dx
y(y+1)
1 1
∫── - ─── dy = ∫dx
y y+1
ln│y│-ln│y+1│ = x + c*
y c* x
── = e * e
y+1
1 -c* -x
1 + ─ = e * e
y
1
y = ───── - 1 令c=e^(-c*)
ce^(-x)
答案一樣~
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