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※ 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~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.162.227.45
topee:感謝 04/07 20:57