Solve (x+y)dy/dx=1 的求解過程?

令 z = x + y , 則 dz = dx + dy
(x+y) dy / dx = 1
(x+y) dy = dx
z ( dz - dx ) = dx
z dz = ( z + 1 ) dx
[ z / (z+1) ] dz = dx

令 u = z + 1 , 則 du = dz
[ ( u - 1 ) / u ] du = dx
[ 1 - 1/u ] du = dx
積分得:
x = u - ln｜u｜+ c1
x = z+1 - ln｜z+1｜+ c1
x = x+y+1 - ln｜x+y+1｜+ c1
y - ln｜x+y+1｜= -1 - c1 ≡ c

Ans: y - ln｜x+y+1｜= c

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驗算:
(1)
當 x+y+1 ≧ 0 , 答案式即為:
y - ln ( x+y+1 ) = c
兩邊對 x 微分得:
y ' - ( 1+ y' )/(x+y+1) = 0
(x+y+1)y' = 1 + y'
(x+y)y' = 1 即 (x+y)dy/dx = 1

(2)
當 x+y+1 < 0 , 答案式即為:
y - ln [ - ( x+y+1 ) ] = c
兩邊對 x 微分得:
y ' - [ - ( 1+ y' )]/[ - (x+y+1) ] = 0
y ' - ( 1+ y' )/(x+y+1) = 0
(x+y+1)y' = 1 + y'
(x+y)y' = 1 即 (x+y)dy/dx = 1
故驗算無誤