令y=n+1,x=m-3,则dy=dn,dx=dm
代入原方程,得dn/dm=(m-n)/(m+4n).(1)
令t=n/m,则dn/dm=mdt/dm+t
代入方程(1),得 mdt/dm=(1-4t^2)/(1+4t)
==> dm/m=(1+4t)dt/(1-4t^2)
==> 2dm/m=[3/(1-2t)-1/(1+2t)]dt
==> 4ln│m│=-3ln│1-2t│-ln│1+2t│+ln│C│ (C是积分常数)
==> m^4=C/[(1+2t)(1-2t)^3]
==> m^4(1+2t)(1-2t)^3=C
==> m^4(1+2n/m)(1-2n/m)^3=C
==> (m-2n)(m+2n)^3=C
==> (x-2y+5)(x+2y+1)^3=C
故 原方程的通解是(x-2y+5)(x+2y+1)^3=C.