常微分方程y'=(x+y)ln(x+y)-1

常微分方程y'=(x+y)ln(x+y)-1
常微分方程y'=(x+y)ln(x+y)-1

常微分方程y'=(x+y)ln(x+y)-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

收起

y'=(x+y)ln(x+y)-1即y'+1=(x+y)ln(x+y).
即(y+x)'=(x+y)ln(x+y)
所以,ln(x+y)=(y+x)'/(y+x)=[ln(x+y)]'
设z=ln(x+y),即z=z',1=z'/z=[ln(z)]'
所以ln(z)=x+C1,即z=e^(x+C1)=Ce^x=ln(x+y)
y=e^[Ce^x]-x (C>0)