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

问题补充：

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

答案：

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

y=xudy=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=xudy=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)

3dy/dt+ytant=sin2t=2sintcost

dy=(2sintcost-ytant)dt

costdy=2sintcostdt+ydcost

(costdy-ydcost)/cost^2=2tantdt

通解y/cost=-2ln|cost|+lnC

4ylnydx+(x-lny)dy=0

y=e^tylny=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

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