dy/dx=(2y+4)/(x+y-1 ) 解这个方程 十万火急如题
问题描述：
dy/dx=(2y+4)/(x+y-1 ) 解这个方程 十万火急
如题
答
x=y²-y/y+4
答
u = y+2
u' = 2u/(x+u-3)
dx/du = 1/(du/dx) = x/2u + 1/2 -3 /2u
dx/du = x/2u => dx/x = du/2u => x = 根号(u) +C
dx/du = 1/2 => x= 0.5u +C
dx/du =-3/2u => x = -1.5 ln|u|
所以x = 根号u + 0.5u -1.5 ln|u| +C
x= 根号(y+2) +0.5 y -1.5 ln|y+2| +C
答
dy/dx=(2y+4)/(x+y-1 )设Y= y+2 X=x-3 代入：dY/dX=2Y/(X+Y)2dX/dY=X/Y+1设U=X/Y X=YU dX/dY=U+YdU/dY 代入：2(U+YdU/dY)=U+12dU/(1-U)=dY/Y积分得：-2ln(1-U)+lnC=lnYY(1-U)²=CY(1-X/Y)²=C(Y-X)²=CY...