Question

求齐次型微分方程的通解

Answer 1

（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).

Answer 2

Answer 3

dy/dx=e^(y/x)+y/x+1
令y/x=u，y=xu
dy/dx=u=xdu/dx
带入原方程得
u+xdu/dx=e^u+u+1
du/(e^u+1)=dx/x
e^u du/e^u (e^u+1)=dx/x
[1/e^u -1/(e^u+1)] d(e^u)=dx/x
lne^u - ln(e^u+1)=ln|x|+ln|C|
e^u/(e^u+1)=Cx
1/[1+e^(-u)]=Cx
e^(-u)=1/Cx -1
-u=ln(1/Cx - 1)
u=-ln(1/Cx - 1)
即y=-x ln(1/Cx - 1)