Question

求一阶线性微分方程的通解

Answer 1

dy/dx=y/(2ylny+y-x)
ydx+(x-y-2ylny)dy=0...........(1)
P=y；Q=x-y-2ylny；∂P/∂y=1=∂Q/∂x；故(1)是全微分方程。
积分之：∫ydx+(x-y-2ylny)dy=∫d(xy-y²lny)=xy-y²lny=C
即xy-y²lny=C为原方程的通解。
代入初始条件，得C=1，故特解为xy-y²lny=1.

追问

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

追问

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)