用这个办法做
追答dx/dy=(2ylny+y-x)/y
=2lny + 1 - x/y
dx/dy + (1/y)x = 2lny + 1
P = 1/y
Q = 2lny + 1
x = Ce^(-∫1/ydy) + e^(-∫1/ydy)∫(2lny+1)e^(∫1/ydy)dy
= Ce^(-lny) + e^(-lny)∫(2lny+1)e^(lny)dy
= C/y + 1/y∫(2lny+1)ydy
= C/y + 1/y(2∫ylnydy+∫ydy)
= C/y + 1/y(∫lnydy^2+y^2/2)
= C/y + 1/y(y^2lny - ∫y^2dlny+y^2/2)
= C/y + 1/y(y^2lny - ∫ydy+y^2/2)
= C/y + ylny
余下的和上面就一样了
用这种方法不是要转换成dy/dx+p(x)y=Q(x)嘛
你怎么是dx/dy+p(x)y=Q(x)