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

1个回答

  • 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