求xdy/dx+1=e^y通解
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∵xdy/dx+1=e^y
==>xdy/dx=e^y-1
==>dy/(e^y-1)=dx/x
==>-e^(-y)dy/(e^(-y)-1)=dx/x
==>d(e^(-y)-1)/(e^(-y)-1)=dx/x
==>ln│e^(-y)-1│=ln│x│+ln│C│ (C是常数)
==>e^(-y)-1=Cx
==>e^(-y)=Cx+1
==>(Cx+1)e^y=1
∴原方程的通解是(Cx+1)e^y=1.
==>xdy/dx=e^y-1
==>dy/(e^y-1)=dx/x
==>-e^(-y)dy/(e^(-y)-1)=dx/x
==>d(e^(-y)-1)/(e^(-y)-1)=dx/x
==>ln│e^(-y)-1│=ln│x│+ln│C│ (C是常数)
==>e^(-y)-1=Cx
==>e^(-y)=Cx+1
==>(Cx+1)e^y=1
∴原方程的通解是(Cx+1)e^y=1.
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