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1/（x+y+1）=y’通解
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1/（x+y+1）=y’通解
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答案和解析
∵1/(x+y+1)=y’
==>dx=(x+y+1)dy
==>dx-xdy=(y+1)dy
==>e^(-y)dx-xe^(-y)dy=(y+1)e^(-y)dy (等式两端同乘e^(-y))
==>d(xe^(-y))=d(-(y+2)e^(-y)) (等式右边应用分部积分法)
==>xe^(-y)=C-(y+2)e^(-y) (C是常数)
==>x=Ce^y-y-2 (等式两端同乘e^y)
∴原方程的通解是x=Ce^y-y-2.
==>dx=(x+y+1)dy
==>dx-xdy=(y+1)dy
==>e^(-y)dx-xe^(-y)dy=(y+1)e^(-y)dy (等式两端同乘e^(-y))
==>d(xe^(-y))=d(-(y+2)e^(-y)) (等式右边应用分部积分法)
==>xe^(-y)=C-(y+2)e^(-y) (C是常数)
==>x=Ce^y-y-2 (等式两端同乘e^y)
∴原方程的通解是x=Ce^y-y-2.
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