Question

常微分方程y'=(x+y)ln(x+y)-1

Answer 1

y'+1=(x+y)ln(x+y)

(dy/dx)+1=(x+y)ln(x+y)

令u=x+y,那么du/dx=1+(dy/dx)

于是上式变为：du/dx=ulnu

于是du/(ulnu)=dx

d(lnu)/lnu=dx

两边同时积分得：ln(lnu)+C1=x+C2

所以通解为：ln[ln(x+y)]-x+C=0