求齐次方程的通解
xt'+t=tlnt==>xt'=t(lnt-1)==>dt/["}}}'>

2个回答

  • (1)令y=xt,则y'=xt'+t

    代入原方程,得y'=(y/x)ln(y/x)

    ==>xt'+t=tlnt

    ==>xt'=t(lnt-1)

    ==>dt/[t(lnt-1)]=dx/x

    ==>d(lnt-1)/(lnt-1)=dx/x

    ==>ln│lnt-1│=ln│x│+ln│C│ (C是积分常数)

    ==>lnt-1=Cx

    ==>lnt=Cx+1

    ==>ln(y/x)=Cx+1

    ==>lny=lnx+Cx+1

    故原方程的通解是lny=lnx+Cx+1 (C是积分常数).

    (2)∵(x²+y²)dx-xydy=0

    ==>(2/x³)(x²+y²)dx=2ydy/x² (等式两端同乘2/x³)

    ==>2ydy/x²-2y²dx/x³=2dx/x

    ==>d(y²/x²)=2dx/x

    ==>y²/x²=ln(x²)+C (C是积分常数)

    ==>y²=x²[ln(x²)+C]

    ∴原方程的通解是y²=x²[ln(x²)+C] (C是积分常数).