求解微分方程dy/dx=(2x+y-1)^2/(x-2)^2
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求解微分方程dy/dx=(2x+y-1)^2/(x-2)^2
令u=x-2,v=y+3,du=dx,dv=dy,
dy/dx=dv/du=((2u+v)/u)^2=4+4v/u+v^2/u^2
Z=v/u,v=zu,dv=udz+zdu,
dv/du=udz/du+z=4+4z+z^2
udz/du=4+3z+z^2
dz/(4+3z+z^2)=du/u
dz/[25/4+(3/2+z)^2]=du/u
两边积分得
2/5arctgan(3/5+2/5z)=lnu+C
2/5arctgan(3/5+2v/5u)=lnu+C
2/5arctgan{3/5+[2(y+3)/5(x-2)]}=lnu+C
dy/dx=dv/du=((2u+v)/u)^2=4+4v/u+v^2/u^2
Z=v/u,v=zu,dv=udz+zdu,
dv/du=udz/du+z=4+4z+z^2
udz/du=4+3z+z^2
dz/(4+3z+z^2)=du/u
dz/[25/4+(3/2+z)^2]=du/u
两边积分得
2/5arctgan(3/5+2/5z)=lnu+C
2/5arctgan(3/5+2v/5u)=lnu+C
2/5arctgan{3/5+[2(y+3)/5(x-2)]}=lnu+C
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