※ 引述《wellilin (QQ12)》之銘言：
: y'+(x+y)x=x^3(x+y)^3-1
: ans: 1/(x+y)^2=(1+x^2)+ce^x^2
3 3
dy + (x+y)xdx = x (x+y) dx - dx
3 3
d(x+y) + x(x+y)dx = x (x+y) dx
白努力~
1-3 -2 -3
令 u = (x+y) = (x+y) , du = -2(x+y) d(x+y)
-1 3 3
---du + xudx = x dx ,du - 2xudx = -2x dx
2
-x^2
I = exp[∫-2xdx] = e
-x^2 3 -x^2
d(ue ) = -2x e dx
-x^2 2 -x^2 2 -v
ue = ∫-x e d(x ) --->看成∫-ve dv
-x^2 -x^2 2
ue = e (x + 1) + c
-2 2 x^2
(x+y) = x + 1 + ce
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