微分方程(x-y+3)dx+(3x+y+1)dy=0的通解

2025-06-21 13:09:07

推荐回答（2个）

回答1：

由x-y+3=0,3x+y+1=0解得x=-1,y=2.
设X=x-1,Y=y+2,原方程变为(X-Y)dX+(3X+Y)dY=0,
设Y=zX,则(1-z)dX+(3+z)(Xdz+zdX)=0,
∴dX/X+(3+z)dz/(1+z)^2=0,
lnX+ln(1+z)-2/(1+z)=C,
∴ln(x-1)+ln[1+(y+2)/(x-1)]-2(x-1)/(x+y+1)=C.
∴ln(x+y+1)-2(x-1)/(x+y+1)=C.

回答2：

解：由x-y+3=0,3x+y+1=0
解得x=-1,y=2.
设X=x-1,Y=y+2,
原方程变为(X-Y)dX+(3X+Y)dY=0,
设Y=zX,则(1-z)dX+(3+z)(Xdz+zdX)=0,
∴dX/X+(3+z)dz/(1+z)^2=0,
lnX+ln(1+z)-2/(1+z)=C,
∴ln(x-1)+ln[1+(y+2)/(x-1)]-2(x-1)/(x+y+1)=C.
∴ln(x+y+1)-2(x-1)/(x+y+1)=C.