Question

y'=(2x-y+1)/(2x-y-1)的通解

Answer 1

设2x-y=t,则y'=2-t'

代入原方程得2-t'=(t+1)/(t-1)

==>t'=(t-3)/(t-1)

==>(t-1)dt/(t-3)=dx

==>[1+2/(t-3)]dt=dx

==>t+2ln│t-3│=x+2ln│C│ (C是积分常数)

==>ln│t-3│=(x-t)/2+ln│C│

==>t-3=Ce^((x-t)/2)

==>2x-y-3=Ce^((y-x)/2)

故原方程的通解是2x-y-3=Ce^((y-x)/2) (C是积分常数).