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解:令t=2-x+y,则y=t+x-2,y'=t'+1
代入原方程,得t'+1=t^2
==>t'=t^2-1
==>dt/(t^2-1)=dx
==>[1/(t-1)-1/(t+1)]dt=2dx
==>ln│t-1│-ln│t+1│=2x+ln│C│ (C是常数)
==>(t-1)/(t+1)=Ce^(2x)
==>t-1=C(t+1)e^(2x)
==>y-x+1=C(3-x+y)e^(2x)
==>y=C(3-x+y)e^(2x)+x-1
故原方程的通解是y=C(3-x+y)e^(2x)+x-1。
代入原方程,得t'+1=t^2
==>t'=t^2-1
==>dt/(t^2-1)=dx
==>[1/(t-1)-1/(t+1)]dt=2dx
==>ln│t-1│-ln│t+1│=2x+ln│C│ (C是常数)
==>(t-1)/(t+1)=Ce^(2x)
==>t-1=C(t+1)e^(2x)
==>y-x+1=C(3-x+y)e^(2x)
==>y=C(3-x+y)e^(2x)+x-1
故原方程的通解是y=C(3-x+y)e^(2x)+x-1。
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