※ 引述《k0184990 (追隨夢想..)》之銘言：
: dy x^2-2x+y^2+4y+5
: ----=------------------
: dx 2(xy+2x-y-2)
: 這題我看了好久才看出來她有以下算法
: dy (x-1)^2+(y+2)^2
: ----=------------------
: dx 2(x-1)(y+2)
: 然後設變數
: 但我想請問除了這方法，還有誰有其他的解法嗎??
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M(x,y) dx + N(x,y) dy = 0 for M(x,y) = x^2 - 2x + y^2 + 4y + 5
N(x,y) = -2(xy + 2x - y - 2)
┌ My = 2y + 4
└ Nx = -2y - 4
假設積分因子 I(x,y)
滿足: (2y + 4)I + M*(Iy) = (-2y - 4)I + N*(Ix)
若 I = I(x) : (2y + 4)I = (-2y - 4)I - 2(xy + 2x - y - 2) * dI/dx
→ 4(y + 2)I = -2(x - 1)(y + 2) * dI/dx
→ 4I = -2(x - 1) * dI/dx or y=-2
解得 I(x) = C/(x-1)^2 (choose C=1)
所以存在一 f(x,y) , 使得 df = M*I dx + N*I dy
→ ┌ fx = (x^2 - 2x + y^2 + 4y + 5)/(x - 1)^2 ____(1)
└ fy = -2(y + 2)/(x - 1) ____(2)
由 (2) 可知 f = -(y^2 + 4y)/(x - 1) + g(x) ____(3), g is a fun. of x
→ fx = (y^2 + 4y)/(x - 1)^2 + g'(x)
比較 (1) 式得 g'(x) = (x^2 - 2x + 5)/(x - 1)^2
= 1 + 4/(x - 1)^2
→ g(x) = x - 4/(x - 1)
因此通解為 f(x,y) = C' , C' is const.
→ -(y^2 + 4y)/(x - 1) + x - 4/(x - 1) = C' by (3)
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ps:
我自己的話，看到有圓錐曲線型態的多項式成分
會先嘗試做座標的平移/旋轉 (也就是原po列的第二條 eq.)
簡化 o.d.e.
再用自己熟悉的方法算
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