Question

请问这个微分方程能用全微分做吗，这道题有几种解法

Answer 1

不用全微分就会比较复杂
(1)
ydx+(x-y)dy=0
ydx+xdy-ydy=0
(ydx+xdy)-ydy=0
因为d(xy)=ydx+xdy
所以原式的全微分为
xy-y²/2=C
或者y=0
(2)
ydx+(x-y)dy=0
y+(x-y)dy/dx=0
两边除x
y/x +(1-y/x)dy/dx=0
设u=y/x
y=xu
dy/dx=u+xdu/dx
则
u+(1-u)(u+xdu/dx)=0
(1-u)(u+xdu/dx)=-u
u+xdu/dx=u/(u-1)
xdu/dx=u/(u-1) -u=(u²-2u)/(1-u)
分离变量
dx/x=(1-u)du/(u²-2u)
因(1-u)/(u²-2u)=(1-u)/[u(u-2)]=(-1/2)[1/u + 1/(u-2)]
即
lnx=(-1/2)[lnu + ln(u-2)]+C1
lnx=(-1/2)[lnu(u-2)]+C1
lnx=ln[1/√u(u-2)]+C1
x=e^[ln[1/√u(u-2)]+C1]
令e^C1=C2
x=C2/√u(u-2)
x√u(u-2)=C2
x²(u²-2u)=C2²
x²(y²/x² -2y/x)=C2²
y²-2xy=C2²
2xy-y²=-C2²
xy-y²/2=-C2²/2
令-C2²/2=C
即xy-y²/2=C

追问

非常感谢