第1个回答 2009-05-23
(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)
y=0
(2)∑ncos(nπ/5)/n!
ncos(npi/5)/n!<=n/n!=1/((n-1)!
n->无穷,可知:1/(n-1)!->0
所以
∑1/(n-1)!是收敛的
由优级数判别法可以知道:
∑ncos(nπ/5)/n!也是收敛的