求微分方程(x^2+2xy)dx+xydy=0的通解

(x² + 2xy)dx + xydy = 0
(1 + 2y/x)dx + y/x dy = 0
令y/x = u，则y = ux，dy = udx + xdu
(1+2u)dx + u²dx + uxdu = 0
(1+u)²dx + xudu = 0
dx/x = -udu/(1+u)²
积分得
lnx = -1/(1+u) - ln(1+u) + C
1/(1+u) + ln[x(1+u)] = C
即x/(x+y) + ln(x+y) = C
这是通解

由n>5时(n-2)(n-3)>(n-1)知
(n-2)!>(n-1)
即(n-1)!>(n-1)²
则1/(n-1)! < 1/(n-1)²
而0 <= |ncos(nπ/5)/n!| <= n/n! = 1/(n-1)! < 1/(n-1)²
且∑1/(n-1)²是收敛的
所以∑ncos(nπ/5)/n!也收敛

(1)
(x² + 2xy)dx + xydy = 0
当x不等于0,y不等于0,两边除以x^2,有:
(1 + 2y/x)dx + y/x dy = 0
令y/x = u(u不等于0)，
则y = ux，dy = udx + xdu
(1+2u)dx + u²dx + uxdu = 0
(1+u)²dx + xudu = 0
dx/x = -udu/(1+u)²
积分得
lnx = -1/(1+u) - ln(1+u) + C
1/(1+u) + ln[x(1+u)] = C
即x/(x+y) + ln(x+y) = C

当u等于0,方程也成立
所以通解是:
x/(x+y)+ln|x+y|=C(C是常数)(y/x不等于0