Re: [理工] 工數微方跟線代

看板Grad-ProbAsk作者doom8199 (～口卡口卡　修～)時間12年前 (2012/01/24 22:44)推噓7(7推 0噓 15→)

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※ 引述《john97611017 (軟哥)》之銘言： : 2. 已知積分因子 u1=xy u2=1/(x^2+y^2) 方程式y'=f(x,y) 求f(x,y) 令 M dx + N dy = 0 ,且 y' = f = -M/N 則由 ┌ (M*u1)_y = (N*u1)_x │ └ (M*u2)_y = (N*u2)_x 可得: ┌ xy(M_y) + xM = xy(N_x) + yN │ │ │ M_y 2y*M N_x 2x*N └ ───── - ─────── = ───── - ─────── x^2 + y^2 (x^2 + y^2)^2 x^2 + y^2 (x^2 + y^2)^2 ┌ xy*(M_y) + x*M = xy*(N_x) + y*N ____(1) => │ └ (x^2+y^2)*(M_y) - 2y*M = (x^2+y^2)*(N_x) - 2x*N ____(2) ┌ (2)+2*(1) ┌ (x+y)^2 *(M_y) + 2(x-y)M = (x+y)^2 *(N_x) - 2(x-y)N => │ => │ └ (2)-2*(1) └ (x-y)^2 *(M_y) - 2(x+y)M = (x-y)^2 *(N_x) - 2(x+y)N ┌ (x+y)^2 *[(M_y) - (N_x)] = -2(x-y)(N+M) ____(3) => │ └ (x-y)^2 *[(M_y) - (N_x)] = -2(x+y)(N-M) ____(4) 兩式相除可得: (x+y)^2 (x-y)*(1-f) ──── = ────── (x-y)^2 (x+y)*(1+f) (x-y)^3 - (x+y)^3 -3(x^2)y - y^3 => f = ───────── = ─────── (x-y)^3 + (x+y)^3 x^3 + 3x(y^2) ------------------------------------------------------------------------------ [驗證] 令 M = [3(x^2)y + y^3]*k(x,y) N = [x^3 + 3x(y^2)]*k(x,y) 分別帶回 (3) (4) 式可得: ┌ (x+y)^2 *{[x^3 + 3x(y^2)]*(k_x)-[3(x^2)y + y^3]*(k_y)} = 2(x-y)(x+y)^3 *k │ └ (x-y)^2 *{[x^3 + 3x(y^2)]*(k_x)-[3(x^2)y + y^3]*(k_y)} = 2(x+y)(x-y)^3 *k => [x^3 + 3x(y^2)]*(k_x) - [3(x^2)y + y^3]*(k_y) = 2(x^2 - y^2)*k 該 pde 的特徵方程式為 : dx -dy dk ─────── = ─────── = ─────── x^3 + 3x(y^2) 3(x^2)y + y^3 2(x^2 - y^2)k ┌ xy(x^2+y^2) = c1 解得 │ └ k = c2(x^2 + y^2) 因此 k(x,y) = (x^2 + y^2)*g( xy(x^2+y^2) ) 即 ┌ M(x,y) = [3(x^2)y + y^3]*(x^2 + y^2)*g( xy(x^2+y^2) ) │ └ N(x,y) = [x^3 + 3x(y^2)]*(x^2 + y^2)*g( xy(x^2+y^2) ) where g(.) is any diff. function of the notation '.' ---- 所以 f 的確滿足題意，只是所選取的 (M,N) 怪噁心就是了.... -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.211.139 ※ 編輯: doom8199 來自: 140.113.211.139 (01/24 22:47)

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