Question

求解下列微分方程 ①dy/dx=(x+y)/(x-y)②（x-y)ydx-x^2dy=0③dy/dt+ytant=sin2t④ylnydx+(x-lny)dy=0

Answer 1

1
dy/dx=(x+y)/(x-y)
y=xu
dy=xdu+udx
xdu+udx=(1+u)/(1-u)dx
xdu=[(1+u)/(1-u)-u]dx
(1-u)du/(1+u^2)=dx/x
arctanu-ln|u|=ln|x|-lnC
arctanu+lnC=ln|y|
y=Ce^(arctan(y/x)通解

2
(xy-y^2)dx-x^2dy=0
y=xu
dy=xdu+udx
(x^2u-x^2u^2)dx-x^2(xdu+udx)=0
(-u^2)dx-xdu=0
dx/x=du/(-u^2)
ln|x|+lnC=1/u
通解Cx=e^(x/y)

3
dy/dt+ytant=sin2t=2sintcost
dy=(2sintcost-ytant)dt
costdy=2sintcostdt+ydcost
(costdy-ydcost)/cost^2=2tantdt
通解y/cost=-2ln|cost|+lnC

4
ylnydx+(x-lny)dy=0
y=e^t
ylny=te^t
te^tdx+(x-t)e^tdt=0
tdx+(x-t)dt=0
tdx+xdt=tdt
xt=(1/2)t^2+C
通解xlny=(1/2)(lny)^2+C