求微分方程dy/dx=（4x+3Y）/(x+y)的通解
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dy/dx=(4x+3y)/(x+y)
dy/dx=3+x/(x+y)
y/x=u dy=udx+xdu
u+xdu/dx=3+1/(1+u)
xdu/dx=3-u+1/(1+u)
(1+u)du/(4+2u-u^2)=dx/x
(-1+u)du/(4+2u-u^2)-2du/(4+2u-u^2)=dx/x
(-1/2)dln(4+2u-u^2)-2du/[5-(u-1)^2]=dlnx
du/[√5-(u-1)][√5+(u-1)]=(1/(2√5))[ln(√5+u-1)-ln(√5-u+1)]
(-1/2)ln(4+2u-u^2)-(1/√5)[ln(√5+u-1)-ln(√5-u+1)]=lnx+C0
(-1/2)ln[4+2y/x-(y/x)^2] - (1/√5)[ln(√5+y/x-1) - ln(√5-y/x+1)]=lnx+C0
dy/dx=3+x/(x+y)
y/x=u dy=udx+xdu
u+xdu/dx=3+1/(1+u)
xdu/dx=3-u+1/(1+u)
(1+u)du/(4+2u-u^2)=dx/x
(-1+u)du/(4+2u-u^2)-2du/(4+2u-u^2)=dx/x
(-1/2)dln(4+2u-u^2)-2du/[5-(u-1)^2]=dlnx
du/[√5-(u-1)][√5+(u-1)]=(1/(2√5))[ln(√5+u-1)-ln(√5-u+1)]
(-1/2)ln(4+2u-u^2)-(1/√5)[ln(√5+u-1)-ln(√5-u+1)]=lnx+C0
(-1/2)ln[4+2y/x-(y/x)^2] - (1/√5)[ln(√5+y/x-1) - ln(√5-y/x+1)]=lnx+C0
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