一階一次常微分方程：齊次方程

代 y = ux, 則 dy = udx + xdu 和 dy/dx = u + x(du/dx)
(1) (x + y)dx - (x + y)dy = 0
dy/dx = 1
y = x + C, 其中 C 為任意常數
(2) xy' - y + √(y2 - x2) = 0
x(u + xu') - ux + x√(u2 - 1) = 0
x(du/dx) = -√(u2 - 1)
du/√(u2 - 1) = - dx/x
∫du/√(u2 - 1) = - ∫dx/x
ln [u + √(u2 - 1)] = - ln x + C

ln [u + √(u2 - 1)] + ln x = C
ln [ux + √(u2x2 - x2)] = C
ux + √(u2x2 - x2) = K
y + √(y2 - x2) = K
√(y2 - x2) = K - y
y2 - x2 = K2 + y2 - 2Ky
2Ky = K2 + x2
y = (K2 + x2)/(2K)

(3) (x2 + y2)dx + x2dy = 0
dy/dx = - [1 + (y/x)2]
u + x(du/dx) = - (1 + u2)
x(du/dx) = - (u2 + u + 1)
du/(u2 + u + 1) = - dx/x
∫du/(u2 + u + 1) = - ∫dx/x
∫du/[(u + 1/2)2 + (√3/2)2] = - ln x + C
(2/√3) tan-1 [(u + 1/2)/(√3/2)] = - ln x + C
tan-1 [(2u + 1)/√3] = (√3/2) (- ln x + C)
(2u + 1)/√3 = tan [(√3/2) (- ln x + C)]
u = (√3/2) tan [(√3/2) (- ln x + C)]
y/x = (√3/2) tan [(√3/2) (- ln x + C)]
y = [(√3/2)x] tan [(√3/2) (- ln x + C)]

2009-09-07 23:00:08 補充：
(1) (x + y)dx - (x - y)dy = 0
dy/dx = [1 + (y/x)]/[1 - (y/x)]
u + xu' = (1 + u)/(1 - u)
xu' = (1 + u^2)/(1 - u)
(1 - u)du/(1 + u^2) = dx/x
∫(1 - u)du/(1 + u^2) = ∫dx/x
∫du/(1 + u^2) - ∫udu/(1 + u^2) = ln x + C

2009-09-07 23:00:20 補充：
tan^-1 u - [ln(1 + u^2)]/2 = ln x + C
tan^-1 u - ln[√(1 + u^2)] = ln x + C
tan^-1 (y/x) - ln[√(1 + y^2/x^2)] = ln x + C

3....
......
(2u + 1)/√3 = tan [(√3/2) (- ln x + C)]
u={√3 tan [(√3/2) (- ln x + C)]-1}/2
y/x={√3 tan [(√3/2) (- ln x + C)]-1}/2
y={√3 tan [(√3/2) (- ln x + C)]-1}x/2

1.打錯咗:( x + y ) d x - ( x - y ) d y = 0