Question

微分方程的解y'=y/(y-x)

Answer 1

解法一：∵y'=y/(y-x) ==>(y-x)y'=y
==>(y-x)dy=ydx
==>ydy=ydx+xdy
==>d(y²)=2d(xy)
==>y²=2xy+C (C是积分常数)
∴原方程的通解是y²=2xy+C (C是积分常数)。
解法二：令y=xt，则y'=xt'+t
代入原方程，得xt'+t=t/(t-1)
==>xt'=t(2-t)/(t-1)
==>(t-1)dt/[t(2-t)]=dx/x
==>[1/(2-t)-1/t]dt=2dx/x
==>ln│t-2│+ln│t│=-2ln│x│+ln│C│ (C是积分常数)
==>t(t-2)=C/x²
==>(y/x)(y/x-2)=C/x²
==>y(y-2x)=C
==>y²-2xy=C
==>y²=2xy+C
故原方程的通解是y²=2xy+C (C是积分常数)。

Answer 2

y' = dy/dx = y/(y - x)
dx/dy = (y - x)/y = 1 - x/y
设 x/y = u，x = yu
dx/dy = u + y du/dy
So, u + y du/dy = 1- u
du/(1 - 2u) = dy/y
- ½ ln|1 - 2u| + c = lny
lny + ½ln|1 - 2u| = c
y²(1 - 2u) = e^c = C
y²(1 - 2x/y) = C
y² - 2xy = C